Does $sumlimits_{n=1}^{infty}frac{cos^{2}(n+1)}{n}$ converge? The Next CEO of Stack...
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Does $sumlimits_{n=1}^{infty}frac{cos^{2}(n+1)}{n}$ converge?
The Next CEO of Stack OverflowDoes the series $sumlimits_{n=1}^{infty}frac{sin(n-sqrt{n^2+n})}{n}$ converge?Does $sumlimits_{n=1}^inftyfrac{1}{sqrt{n}+sqrt{n+1}}$ converge?Is $sum cos(n pi) frac{n}{n^2+1}$ conditionally or absolutely convergent?Does $sumlimits_{n=1}^inftysin(n)sinleft(frac{pi}{2n}right)$ converge?Alternating series: $sumlimits_{n= 1}^{infty} (-1)^{n-1} frac{ln(n)}{n}$ convergence?On the convergence of $sum_{n=1}^{+infty}frac{1}{n},cosleft(frac{pi n}{2}right)$Does $sum _{n=1}^{infty }left(-1right)^{n+1}left(1-cosleft(frac{1}{sqrt{n}}right)right)$ converges conditionally?$sumlimits_{n=1}^{infty} frac{tan^{-1} n}{n}$ diverges.Does the series $sum_{n=1}^infty frac{n}{sqrt[3]{8n^5-1}}$ Converge?Convergence of $sumlimits_{m= 0}^{infty} frac{(-3)^m+2^m}{3^m+2^m}$
$begingroup$
The original question, given to my Calculus II recitation class, was:
Determine if the series $$sumlimits_{n=1}^{infty}frac{(-1)^{n}cos^{2}(n+1)}{n}$$
converges absolutely, conditionally, or diverges. I can kind of see a comparison with the alternating harmonic series here, but making that formal is tough. With the absolute series $sumlimits_{n=1}^{infty} frac{cos^{2}(n+1)}{n}$, I'm not sure what test to apply.
What I've Tried: No tests (in the classical Calc II curriculum) work. I've tried expanding $cos^{2}(n+1)$ into a power series within the series in question, but I'm not really sure where to go from there. My intuition tells me this series will diverge, since it seems "close" to the harmonic series; but $cos(x)$ is less than $1$ infinitely often.
real-analysis calculus sequences-and-series trigonometry
edited Mar 17 at 8:46
rtybase
11.5k31534
asked May 6 '18 at 5:21
FredFred
62659
$endgroup$
add a comment |
$begingroup$
The original question, given to my Calculus II recitation class, was:
Determine if the series $$sumlimits_{n=1}^{infty}frac{(-1)^{n}cos^{2}(n+1)}{n}$$
converges absolutely, conditionally, or diverges. I can kind of see a comparison with the alternating harmonic series here, but making that formal is tough. With the absolute series $sumlimits_{n=1}^{infty} frac{cos^{2}(n+1)}{n}$, I'm not sure what test to apply.
What I've Tried: No tests (in the classical Calc II curriculum) work. I've tried expanding $cos^{2}(n+1)$ into a power series within the series in question, but I'm not really sure where to go from there. My intuition tells me this series will diverge, since it seems "close" to the harmonic series; but $cos(x)$ is less than $1$ infinitely often.
real-analysis calculus sequences-and-series trigonometry
edited Mar 17 at 8:46
rtybase
11.5k31534
asked May 6 '18 at 5:21
FredFred
62659
$endgroup$
$begingroup$
Fred, if this question is from a Calculus II course, my question is, how are you supposed to come up with an answer using the techniques available from Calculus II material? You may want to convey that to your professor. Which brings me to the second question. From which textbook is this exercise? Just curious
$endgroup$
– imranfat
May 6 '18 at 5:49
$begingroup$
@imranfat This is from a practice sheet made by the instructor. I'm his TA for the course. Doing this on the board, I didn't immediately know what to do; and thinking about it on and off, I'm convinced it's a mistake, as I don't see any immediate way to apply the traditional Calculus II material. Regardless, it's an interesting question that's been bugging me.
$endgroup$
– Fred
May 6 '18 at 5:52
$begingroup$
Well, you can ask your professor what test he expects the students doe carry out. He doesn't have to show the work; you can share that with us. I am just curious...
$endgroup$
– imranfat
May 6 '18 at 16:11
add a comment |
$begingroup$
The original question, given to my Calculus II recitation class, was:
Determine if the series $$sumlimits_{n=1}^{infty}frac{(-1)^{n}cos^{2}(n+1)}{n}$$
converges absolutely, conditionally, or diverges. I can kind of see a comparison with the alternating harmonic series here, but making that formal is tough. With the absolute series $sumlimits_{n=1}^{infty} frac{cos^{2}(n+1)}{n}$, I'm not sure what test to apply.
What I've Tried: No tests (in the classical Calc II curriculum) work. I've tried expanding $cos^{2}(n+1)$ into a power series within the series in question, but I'm not really sure where to go from there. My intuition tells me this series will diverge, since it seems "close" to the harmonic series; but $cos(x)$ is less than $1$ infinitely often.
real-analysis calculus sequences-and-series trigonometry
edited Mar 17 at 8:46
rtybase
11.5k31534
asked May 6 '18 at 5:21
FredFred
62659
$endgroup$
The original question, given to my Calculus II recitation class, was:
Determine if the series $$sumlimits_{n=1}^{infty}frac{(-1)^{n}cos^{2}(n+1)}{n}$$
converges absolutely, conditionally, or diverges. I can kind of see a comparison with the alternating harmonic series here, but making that formal is tough. With the absolute series $sumlimits_{n=1}^{infty} frac{cos^{2}(n+1)}{n}$, I'm not sure what test to apply.
What I've Tried: No tests (in the classical Calc II curriculum) work. I've tried expanding $cos^{2}(n+1)$ into a power series within the series in question, but I'm not really sure where to go from there. My intuition tells me this series will diverge, since it seems "close" to the harmonic series; but $cos(x)$ is less than $1$ infinitely often.
real-analysis calculus sequences-and-series trigonometry
real-analysis calculus sequences-and-series trigonometry
edited Mar 17 at 8:46
rtybase
11.5k31534
asked May 6 '18 at 5:21
FredFred
62659
edited Mar 17 at 8:46
rtybase
11.5k31534
asked May 6 '18 at 5:21
FredFred
62659
edited Mar 17 at 8:46
rtybase
11.5k31534
edited Mar 17 at 8:46
rtybase
11.5k31534
edited Mar 17 at 8:46
rtybase
11.5k31534
11.5k31534
asked May 6 '18 at 5:21
FredFred
62659
asked May 6 '18 at 5:21
FredFred
62659
asked May 6 '18 at 5:21
FredFred
62659
62659
$begingroup$
Fred, if this question is from a Calculus II course, my question is, how are you supposed to come up with an answer using the techniques available from Calculus II material? You may want to convey that to your professor. Which brings me to the second question. From which textbook is this exercise? Just curious
$endgroup$
– imranfat
May 6 '18 at 5:49
$begingroup$
@imranfat This is from a practice sheet made by the instructor. I'm his TA for the course. Doing this on the board, I didn't immediately know what to do; and thinking about it on and off, I'm convinced it's a mistake, as I don't see any immediate way to apply the traditional Calculus II material. Regardless, it's an interesting question that's been bugging me.
$endgroup$
– Fred
May 6 '18 at 5:52
$begingroup$
Well, you can ask your professor what test he expects the students doe carry out. He doesn't have to show the work; you can share that with us. I am just curious...
$endgroup$
– imranfat
May 6 '18 at 16:11
add a comment |
$begingroup$
Fred, if this question is from a Calculus II course, my question is, how are you supposed to come up with an answer using the techniques available from Calculus II material? You may want to convey that to your professor. Which brings me to the second question. From which textbook is this exercise? Just curious
$endgroup$
– imranfat
May 6 '18 at 5:49
$begingroup$
@imranfat This is from a practice sheet made by the instructor. I'm his TA for the course. Doing this on the board, I didn't immediately know what to do; and thinking about it on and off, I'm convinced it's a mistake, as I don't see any immediate way to apply the traditional Calculus II material. Regardless, it's an interesting question that's been bugging me.
$endgroup$
– Fred
May 6 '18 at 5:52
$begingroup$
Well, you can ask your professor what test he expects the students doe carry out. He doesn't have to show the work; you can share that with us. I am just curious...
$endgroup$
– imranfat
May 6 '18 at 16:11
$begingroup$
Fred, if this question is from a Calculus II course, my question is, how are you supposed to come up with an answer using the techniques available from Calculus II material? You may want to convey that to your professor. Which brings me to the second question. From which textbook is this exercise? Just curious
$endgroup$
– imranfat
May 6 '18 at 5:49
$begingroup$
Fred, if this question is from a Calculus II course, my question is, how are you supposed to come up with an answer using the techniques available from Calculus II material? You may want to convey that to your professor. Which brings me to the second question. From which textbook is this exercise? Just curious
$endgroup$
– imranfat
May 6 '18 at 5:49
$begingroup$
@imranfat This is from a practice sheet made by the instructor. I'm his TA for the course. Doing this on the board, I didn't immediately know what to do; and thinking about it on and off, I'm convinced it's a mistake, as I don't see any immediate way to apply the traditional Calculus II material. Regardless, it's an interesting question that's been bugging me.
$endgroup$
– Fred
May 6 '18 at 5:52
$begingroup$
@imranfat This is from a practice sheet made by the instructor. I'm his TA for the course. Doing this on the board, I didn't immediately know what to do; and thinking about it on and off, I'm convinced it's a mistake, as I don't see any immediate way to apply the traditional Calculus II material. Regardless, it's an interesting question that's been bugging me.
$endgroup$
– Fred
May 6 '18 at 5:52
$begingroup$
Well, you can ask your professor what test he expects the students doe carry out. He doesn't have to show the work; you can share that with us. I am just curious...
$endgroup$
– imranfat
May 6 '18 at 16:11
$begingroup$
Well, you can ask your professor what test he expects the students doe carry out. He doesn't have to show the work; you can share that with us. I am just curious...
$endgroup$
– imranfat
May 6 '18 at 16:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}$ clearly diverges. For any two consecutive integers, at least one of them is distant from a (non-integer) half-integer multiple of $pi$ by at least $frac{1}{2}$, which is more than $frac{pi}{8}$. That number of the pair will have a cosine greater than $cos(frac{3pi}{8})$ in absolute value, and $cos(frac{3pi}{8})=sin(frac{pi}{8})>frac{1}{2}sin(frac{pi}{4})=frac{sqrt{2}}{4}>frac{1}{3}$. Thus, any two consecutive terms $frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1}$ will contribute at least $frac{1}{9(j+1)}$ to the sum. Since $jge 1$, we have $frac{1}{j+1}gefrac{1}{4j}+frac{1}{4(j+1)}$, so $frac{1}{9(j+1)} gefrac{1}{36j}+frac{1}{36(j+1)}$, i.e. $$frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1} ge frac{1}{36j}+frac{1}{36(j+1)}$$ But this means $$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}gefrac{1}{36}sum_{n=1}^{infty}frac{1}{n}$$ which diverges.
answered May 6 '18 at 5:43
C MonsourC Monsour
6,3391326
$endgroup$
add a comment |
$begingroup$
Regarding the question on absolute convergence, the answer is no.
$$sum_{n=1}^mfrac{cos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{1}{2n}}_{text{divergent harmonic series}} + underbrace{sum_{n=1}^mfrac{cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Also we can use the same approach to prove convergence when $(-1)^n$ appears.
$$sum_{n=1}^mfrac{(-1)^ncos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{(-1)^n}{2n}}_{text{convergent alternating series}} - underbrace{sum_{n=1}^mfrac{(-1)^{n+1}cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Note that $(-1)^{n+1}cos 2(n+1) = cos [(n+1)pi] cos 2(n+1) = cos [(n+1)(2 + pi)]$
answered May 6 '18 at 5:35
RRLRRL
53.1k52574
$endgroup$
$begingroup$
Thanks. I see you've distributed summation. Is this legal in this case? Also, I've wiki'd Dirichlet's test, but I can't see which two series you are using to fit the criterion.
$endgroup$
– Fred
May 6 '18 at 5:41
1
$begingroup$
@Fred. Yes you can split the partial sums as I added above. As $m to infty$ the first sum on the RHS diverges and the second sum converges so the sum on the LHS diverges.
$endgroup$
– RRL
May 6 '18 at 5:45
2
$begingroup$
This is valid but difficult. (In particular, I think a Calc II student might have trouble finding a bound for $sum cos 2(n+1)$ so even if Dirichlet's test was part of his curriculum it might be tough. I give a more direct and elementary argument below.
$endgroup$
– C Monsour
May 6 '18 at 5:49
$begingroup$
@Fred: I think you figured it out. The Dirichlet test applies because $1/(2n) to 0$ in a decreasing fashion and the partial sums $sum_{n=1}^{m} cos 2(n+1)$ are bounded for all $m$. Same applies to $sum_{n=1}^m cos ((2+pi)(n+1)$.
$endgroup$
– RRL
May 6 '18 at 6:06
add a comment |
$begingroup$
You could use Dirichlet's test. Observe $n^{-1}$ converges monotonically to zero and
begin{align}
left|sum^N_{n=1}(-1)^n cos^2(n+1) right|leq 1.
end{align}
Hence the series converges.
answered May 6 '18 at 5:35
Jacky ChongJacky Chong
19.3k21129
$endgroup$
1
$begingroup$
True, but this test is not in Calculus II curriculum.
$endgroup$
– imranfat
May 6 '18 at 5:45
$begingroup$
@CMonsour Yes. But I didn't say $sum^N_{n=1}cos^2(n+1)$ is bounded. Look carefully.
$endgroup$
– Jacky Chong
May 6 '18 at 5:48
$begingroup$
OK, I see what your claim is now. (I had missed the $(-1)^n$ since the question title was about the absolute series.) It does seem like it might be a difficult bound for a Calc II student to prove.
$endgroup$
– C Monsour
May 6 '18 at 5:55
$begingroup$
Jacky.Very nice, clear and concise.Thanks!
$endgroup$
– Peter Szilas
May 6 '18 at 6:51
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
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$begingroup$
$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}$ clearly diverges. For any two consecutive integers, at least one of them is distant from a (non-integer) half-integer multiple of $pi$ by at least $frac{1}{2}$, which is more than $frac{pi}{8}$. That number of the pair will have a cosine greater than $cos(frac{3pi}{8})$ in absolute value, and $cos(frac{3pi}{8})=sin(frac{pi}{8})>frac{1}{2}sin(frac{pi}{4})=frac{sqrt{2}}{4}>frac{1}{3}$. Thus, any two consecutive terms $frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1}$ will contribute at least $frac{1}{9(j+1)}$ to the sum. Since $jge 1$, we have $frac{1}{j+1}gefrac{1}{4j}+frac{1}{4(j+1)}$, so $frac{1}{9(j+1)} gefrac{1}{36j}+frac{1}{36(j+1)}$, i.e. $$frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1} ge frac{1}{36j}+frac{1}{36(j+1)}$$ But this means $$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}gefrac{1}{36}sum_{n=1}^{infty}frac{1}{n}$$ which diverges.
answered May 6 '18 at 5:43
C MonsourC Monsour
6,3391326
$endgroup$
add a comment |
$begingroup$
$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}$ clearly diverges. For any two consecutive integers, at least one of them is distant from a (non-integer) half-integer multiple of $pi$ by at least $frac{1}{2}$, which is more than $frac{pi}{8}$. That number of the pair will have a cosine greater than $cos(frac{3pi}{8})$ in absolute value, and $cos(frac{3pi}{8})=sin(frac{pi}{8})>frac{1}{2}sin(frac{pi}{4})=frac{sqrt{2}}{4}>frac{1}{3}$. Thus, any two consecutive terms $frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1}$ will contribute at least $frac{1}{9(j+1)}$ to the sum. Since $jge 1$, we have $frac{1}{j+1}gefrac{1}{4j}+frac{1}{4(j+1)}$, so $frac{1}{9(j+1)} gefrac{1}{36j}+frac{1}{36(j+1)}$, i.e. $$frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1} ge frac{1}{36j}+frac{1}{36(j+1)}$$ But this means $$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}gefrac{1}{36}sum_{n=1}^{infty}frac{1}{n}$$ which diverges.
answered May 6 '18 at 5:43
C MonsourC Monsour
6,3391326
$endgroup$
add a comment |
$begingroup$
$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}$ clearly diverges. For any two consecutive integers, at least one of them is distant from a (non-integer) half-integer multiple of $pi$ by at least $frac{1}{2}$, which is more than $frac{pi}{8}$. That number of the pair will have a cosine greater than $cos(frac{3pi}{8})$ in absolute value, and $cos(frac{3pi}{8})=sin(frac{pi}{8})>frac{1}{2}sin(frac{pi}{4})=frac{sqrt{2}}{4}>frac{1}{3}$. Thus, any two consecutive terms $frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1}$ will contribute at least $frac{1}{9(j+1)}$ to the sum. Since $jge 1$, we have $frac{1}{j+1}gefrac{1}{4j}+frac{1}{4(j+1)}$, so $frac{1}{9(j+1)} gefrac{1}{36j}+frac{1}{36(j+1)}$, i.e. $$frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1} ge frac{1}{36j}+frac{1}{36(j+1)}$$ But this means $$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}gefrac{1}{36}sum_{n=1}^{infty}frac{1}{n}$$ which diverges.
answered May 6 '18 at 5:43
C MonsourC Monsour
6,3391326
$endgroup$
$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}$ clearly diverges. For any two consecutive integers, at least one of them is distant from a (non-integer) half-integer multiple of $pi$ by at least $frac{1}{2}$, which is more than $frac{pi}{8}$. That number of the pair will have a cosine greater than $cos(frac{3pi}{8})$ in absolute value, and $cos(frac{3pi}{8})=sin(frac{pi}{8})>frac{1}{2}sin(frac{pi}{4})=frac{sqrt{2}}{4}>frac{1}{3}$. Thus, any two consecutive terms $frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1}$ will contribute at least $frac{1}{9(j+1)}$ to the sum. Since $jge 1$, we have $frac{1}{j+1}gefrac{1}{4j}+frac{1}{4(j+1)}$, so $frac{1}{9(j+1)} gefrac{1}{36j}+frac{1}{36(j+1)}$, i.e. $$frac{cos^{2}(j+1)}{j} + frac{cos^{2}((j+1)+1)}{j+1} ge frac{1}{36j}+frac{1}{36(j+1)}$$ But this means $$sum_{n=1}^{infty}frac{cos^{2}(n+1)}{n}gefrac{1}{36}sum_{n=1}^{infty}frac{1}{n}$$ which diverges.
answered May 6 '18 at 5:43
C MonsourC Monsour
6,3391326
edited Mar 17 at 4:12
answered May 6 '18 at 5:43
C MonsourC Monsour
6,3391326
answered May 6 '18 at 5:43
C MonsourC Monsour
6,3391326
answered May 6 '18 at 5:43
C MonsourC Monsour
6,3391326
6,3391326
add a comment |
add a comment |
$begingroup$
Regarding the question on absolute convergence, the answer is no.
$$sum_{n=1}^mfrac{cos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{1}{2n}}_{text{divergent harmonic series}} + underbrace{sum_{n=1}^mfrac{cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Also we can use the same approach to prove convergence when $(-1)^n$ appears.
$$sum_{n=1}^mfrac{(-1)^ncos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{(-1)^n}{2n}}_{text{convergent alternating series}} - underbrace{sum_{n=1}^mfrac{(-1)^{n+1}cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Note that $(-1)^{n+1}cos 2(n+1) = cos [(n+1)pi] cos 2(n+1) = cos [(n+1)(2 + pi)]$
answered May 6 '18 at 5:35
RRLRRL
53.1k52574
$endgroup$
$begingroup$
Thanks. I see you've distributed summation. Is this legal in this case? Also, I've wiki'd Dirichlet's test, but I can't see which two series you are using to fit the criterion.
$endgroup$
– Fred
May 6 '18 at 5:41
1
$begingroup$
@Fred. Yes you can split the partial sums as I added above. As $m to infty$ the first sum on the RHS diverges and the second sum converges so the sum on the LHS diverges.
$endgroup$
– RRL
May 6 '18 at 5:45
2
$begingroup$
This is valid but difficult. (In particular, I think a Calc II student might have trouble finding a bound for $sum cos 2(n+1)$ so even if Dirichlet's test was part of his curriculum it might be tough. I give a more direct and elementary argument below.
$endgroup$
– C Monsour
May 6 '18 at 5:49
$begingroup$
@Fred: I think you figured it out. The Dirichlet test applies because $1/(2n) to 0$ in a decreasing fashion and the partial sums $sum_{n=1}^{m} cos 2(n+1)$ are bounded for all $m$. Same applies to $sum_{n=1}^m cos ((2+pi)(n+1)$.
$endgroup$
– RRL
May 6 '18 at 6:06
add a comment |
$begingroup$
Regarding the question on absolute convergence, the answer is no.
$$sum_{n=1}^mfrac{cos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{1}{2n}}_{text{divergent harmonic series}} + underbrace{sum_{n=1}^mfrac{cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Also we can use the same approach to prove convergence when $(-1)^n$ appears.
$$sum_{n=1}^mfrac{(-1)^ncos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{(-1)^n}{2n}}_{text{convergent alternating series}} - underbrace{sum_{n=1}^mfrac{(-1)^{n+1}cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Note that $(-1)^{n+1}cos 2(n+1) = cos [(n+1)pi] cos 2(n+1) = cos [(n+1)(2 + pi)]$
answered May 6 '18 at 5:35
RRLRRL
53.1k52574
$endgroup$
$begingroup$
Thanks. I see you've distributed summation. Is this legal in this case? Also, I've wiki'd Dirichlet's test, but I can't see which two series you are using to fit the criterion.
$endgroup$
– Fred
May 6 '18 at 5:41
1
$begingroup$
@Fred. Yes you can split the partial sums as I added above. As $m to infty$ the first sum on the RHS diverges and the second sum converges so the sum on the LHS diverges.
$endgroup$
– RRL
May 6 '18 at 5:45
2
$begingroup$
This is valid but difficult. (In particular, I think a Calc II student might have trouble finding a bound for $sum cos 2(n+1)$ so even if Dirichlet's test was part of his curriculum it might be tough. I give a more direct and elementary argument below.
$endgroup$
– C Monsour
May 6 '18 at 5:49
$begingroup$
@Fred: I think you figured it out. The Dirichlet test applies because $1/(2n) to 0$ in a decreasing fashion and the partial sums $sum_{n=1}^{m} cos 2(n+1)$ are bounded for all $m$. Same applies to $sum_{n=1}^m cos ((2+pi)(n+1)$.
$endgroup$
– RRL
May 6 '18 at 6:06
add a comment |
$begingroup$
Regarding the question on absolute convergence, the answer is no.
$$sum_{n=1}^mfrac{cos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{1}{2n}}_{text{divergent harmonic series}} + underbrace{sum_{n=1}^mfrac{cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Also we can use the same approach to prove convergence when $(-1)^n$ appears.
$$sum_{n=1}^mfrac{(-1)^ncos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{(-1)^n}{2n}}_{text{convergent alternating series}} - underbrace{sum_{n=1}^mfrac{(-1)^{n+1}cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Note that $(-1)^{n+1}cos 2(n+1) = cos [(n+1)pi] cos 2(n+1) = cos [(n+1)(2 + pi)]$
answered May 6 '18 at 5:35
RRLRRL
53.1k52574
$endgroup$
Regarding the question on absolute convergence, the answer is no.
$$sum_{n=1}^mfrac{cos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{1}{2n}}_{text{divergent harmonic series}} + underbrace{sum_{n=1}^mfrac{cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Also we can use the same approach to prove convergence when $(-1)^n$ appears.
$$sum_{n=1}^mfrac{(-1)^ncos^2 (n+1)}{n} = underbrace{sum_{n=1}^mfrac{(-1)^n}{2n}}_{text{convergent alternating series}} - underbrace{sum_{n=1}^mfrac{(-1)^{n+1}cos 2(n+1)}{2n}}_{text{convergent by Dirichlet test}}$$
Note that $(-1)^{n+1}cos 2(n+1) = cos [(n+1)pi] cos 2(n+1) = cos [(n+1)(2 + pi)]$
answered May 6 '18 at 5:35
RRLRRL
53.1k52574
edited May 6 '18 at 5:53
answered May 6 '18 at 5:35
RRLRRL
53.1k52574
answered May 6 '18 at 5:35
RRLRRL
53.1k52574
answered May 6 '18 at 5:35
RRLRRL
53.1k52574
53.1k52574
$begingroup$
Thanks. I see you've distributed summation. Is this legal in this case? Also, I've wiki'd Dirichlet's test, but I can't see which two series you are using to fit the criterion.
$endgroup$
– Fred
May 6 '18 at 5:41
1
$begingroup$
@Fred. Yes you can split the partial sums as I added above. As $m to infty$ the first sum on the RHS diverges and the second sum converges so the sum on the LHS diverges.
$endgroup$
– RRL
May 6 '18 at 5:45
2
$begingroup$
This is valid but difficult. (In particular, I think a Calc II student might have trouble finding a bound for $sum cos 2(n+1)$ so even if Dirichlet's test was part of his curriculum it might be tough. I give a more direct and elementary argument below.
$endgroup$
– C Monsour
May 6 '18 at 5:49
$begingroup$
@Fred: I think you figured it out. The Dirichlet test applies because $1/(2n) to 0$ in a decreasing fashion and the partial sums $sum_{n=1}^{m} cos 2(n+1)$ are bounded for all $m$. Same applies to $sum_{n=1}^m cos ((2+pi)(n+1)$.
$endgroup$
– RRL
May 6 '18 at 6:06
add a comment |
$begingroup$
Thanks. I see you've distributed summation. Is this legal in this case? Also, I've wiki'd Dirichlet's test, but I can't see which two series you are using to fit the criterion.
$endgroup$
– Fred
May 6 '18 at 5:41
1
$begingroup$
@Fred. Yes you can split the partial sums as I added above. As $m to infty$ the first sum on the RHS diverges and the second sum converges so the sum on the LHS diverges.
$endgroup$
– RRL
May 6 '18 at 5:45
2
$begingroup$
This is valid but difficult. (In particular, I think a Calc II student might have trouble finding a bound for $sum cos 2(n+1)$ so even if Dirichlet's test was part of his curriculum it might be tough. I give a more direct and elementary argument below.
$endgroup$
– C Monsour
May 6 '18 at 5:49
$begingroup$
@Fred: I think you figured it out. The Dirichlet test applies because $1/(2n) to 0$ in a decreasing fashion and the partial sums $sum_{n=1}^{m} cos 2(n+1)$ are bounded for all $m$. Same applies to $sum_{n=1}^m cos ((2+pi)(n+1)$.
$endgroup$
– RRL
May 6 '18 at 6:06
$begingroup$
Thanks. I see you've distributed summation. Is this legal in this case? Also, I've wiki'd Dirichlet's test, but I can't see which two series you are using to fit the criterion.
$endgroup$
– Fred
May 6 '18 at 5:41
$begingroup$
Thanks. I see you've distributed summation. Is this legal in this case? Also, I've wiki'd Dirichlet's test, but I can't see which two series you are using to fit the criterion.
$endgroup$
– Fred
May 6 '18 at 5:41
1
1
$begingroup$
@Fred. Yes you can split the partial sums as I added above. As $m to infty$ the first sum on the RHS diverges and the second sum converges so the sum on the LHS diverges.
$endgroup$
– RRL
May 6 '18 at 5:45
$begingroup$
@Fred. Yes you can split the partial sums as I added above. As $m to infty$ the first sum on the RHS diverges and the second sum converges so the sum on the LHS diverges.
$endgroup$
– RRL
May 6 '18 at 5:45
2
2
$begingroup$
This is valid but difficult. (In particular, I think a Calc II student might have trouble finding a bound for $sum cos 2(n+1)$ so even if Dirichlet's test was part of his curriculum it might be tough. I give a more direct and elementary argument below.
$endgroup$
– C Monsour
May 6 '18 at 5:49
$begingroup$
This is valid but difficult. (In particular, I think a Calc II student might have trouble finding a bound for $sum cos 2(n+1)$ so even if Dirichlet's test was part of his curriculum it might be tough. I give a more direct and elementary argument below.
$endgroup$
– C Monsour
May 6 '18 at 5:49
$begingroup$
@Fred: I think you figured it out. The Dirichlet test applies because $1/(2n) to 0$ in a decreasing fashion and the partial sums $sum_{n=1}^{m} cos 2(n+1)$ are bounded for all $m$. Same applies to $sum_{n=1}^m cos ((2+pi)(n+1)$.
$endgroup$
– RRL
May 6 '18 at 6:06
$begingroup$
@Fred: I think you figured it out. The Dirichlet test applies because $1/(2n) to 0$ in a decreasing fashion and the partial sums $sum_{n=1}^{m} cos 2(n+1)$ are bounded for all $m$. Same applies to $sum_{n=1}^m cos ((2+pi)(n+1)$.
$endgroup$
– RRL
May 6 '18 at 6:06
add a comment |
$begingroup$
You could use Dirichlet's test. Observe $n^{-1}$ converges monotonically to zero and
begin{align}
left|sum^N_{n=1}(-1)^n cos^2(n+1) right|leq 1.
end{align}
Hence the series converges.
answered May 6 '18 at 5:35
Jacky ChongJacky Chong
19.3k21129
$endgroup$
1
$begingroup$
True, but this test is not in Calculus II curriculum.
$endgroup$
– imranfat
May 6 '18 at 5:45
$begingroup$
@CMonsour Yes. But I didn't say $sum^N_{n=1}cos^2(n+1)$ is bounded. Look carefully.
$endgroup$
– Jacky Chong
May 6 '18 at 5:48
$begingroup$
OK, I see what your claim is now. (I had missed the $(-1)^n$ since the question title was about the absolute series.) It does seem like it might be a difficult bound for a Calc II student to prove.
$endgroup$
– C Monsour
May 6 '18 at 5:55
$begingroup$
Jacky.Very nice, clear and concise.Thanks!
$endgroup$
– Peter Szilas
May 6 '18 at 6:51
add a comment |
$begingroup$
You could use Dirichlet's test. Observe $n^{-1}$ converges monotonically to zero and
begin{align}
left|sum^N_{n=1}(-1)^n cos^2(n+1) right|leq 1.
end{align}
Hence the series converges.
answered May 6 '18 at 5:35
Jacky ChongJacky Chong
19.3k21129
$endgroup$
1
$begingroup$
True, but this test is not in Calculus II curriculum.
$endgroup$
– imranfat
May 6 '18 at 5:45
$begingroup$
@CMonsour Yes. But I didn't say $sum^N_{n=1}cos^2(n+1)$ is bounded. Look carefully.
$endgroup$
– Jacky Chong
May 6 '18 at 5:48
$begingroup$
OK, I see what your claim is now. (I had missed the $(-1)^n$ since the question title was about the absolute series.) It does seem like it might be a difficult bound for a Calc II student to prove.
$endgroup$
– C Monsour
May 6 '18 at 5:55
$begingroup$
Jacky.Very nice, clear and concise.Thanks!
$endgroup$
– Peter Szilas
May 6 '18 at 6:51
add a comment |
$begingroup$
You could use Dirichlet's test. Observe $n^{-1}$ converges monotonically to zero and
begin{align}
left|sum^N_{n=1}(-1)^n cos^2(n+1) right|leq 1.
end{align}
Hence the series converges.
answered May 6 '18 at 5:35
Jacky ChongJacky Chong
19.3k21129
$endgroup$
You could use Dirichlet's test. Observe $n^{-1}$ converges monotonically to zero and
begin{align}
left|sum^N_{n=1}(-1)^n cos^2(n+1) right|leq 1.
end{align}
Hence the series converges.
answered May 6 '18 at 5:35
Jacky ChongJacky Chong
19.3k21129
answered May 6 '18 at 5:35
Jacky ChongJacky Chong
19.3k21129
answered May 6 '18 at 5:35
Jacky ChongJacky Chong
19.3k21129
answered May 6 '18 at 5:35
Jacky ChongJacky Chong
19.3k21129
19.3k21129
1
$begingroup$
True, but this test is not in Calculus II curriculum.
$endgroup$
– imranfat
May 6 '18 at 5:45
$begingroup$
@CMonsour Yes. But I didn't say $sum^N_{n=1}cos^2(n+1)$ is bounded. Look carefully.
$endgroup$
– Jacky Chong
May 6 '18 at 5:48
$begingroup$
OK, I see what your claim is now. (I had missed the $(-1)^n$ since the question title was about the absolute series.) It does seem like it might be a difficult bound for a Calc II student to prove.
$endgroup$
– C Monsour
May 6 '18 at 5:55
$begingroup$
Jacky.Very nice, clear and concise.Thanks!
$endgroup$
– Peter Szilas
May 6 '18 at 6:51
add a comment |
1
$begingroup$
True, but this test is not in Calculus II curriculum.
$endgroup$
– imranfat
May 6 '18 at 5:45
$begingroup$
@CMonsour Yes. But I didn't say $sum^N_{n=1}cos^2(n+1)$ is bounded. Look carefully.
$endgroup$
– Jacky Chong
May 6 '18 at 5:48
$begingroup$
OK, I see what your claim is now. (I had missed the $(-1)^n$ since the question title was about the absolute series.) It does seem like it might be a difficult bound for a Calc II student to prove.
$endgroup$
– C Monsour
May 6 '18 at 5:55
$begingroup$
Jacky.Very nice, clear and concise.Thanks!
$endgroup$
– Peter Szilas
May 6 '18 at 6:51
1
1
$begingroup$
True, but this test is not in Calculus II curriculum.
$endgroup$
– imranfat
May 6 '18 at 5:45
$begingroup$
True, but this test is not in Calculus II curriculum.
$endgroup$
– imranfat
May 6 '18 at 5:45
$begingroup$
@CMonsour Yes. But I didn't say $sum^N_{n=1}cos^2(n+1)$ is bounded. Look carefully.
$endgroup$
– Jacky Chong
May 6 '18 at 5:48
$begingroup$
@CMonsour Yes. But I didn't say $sum^N_{n=1}cos^2(n+1)$ is bounded. Look carefully.
$endgroup$
– Jacky Chong
May 6 '18 at 5:48
$begingroup$
OK, I see what your claim is now. (I had missed the $(-1)^n$ since the question title was about the absolute series.) It does seem like it might be a difficult bound for a Calc II student to prove.
$endgroup$
– C Monsour
May 6 '18 at 5:55
$begingroup$
OK, I see what your claim is now. (I had missed the $(-1)^n$ since the question title was about the absolute series.) It does seem like it might be a difficult bound for a Calc II student to prove.
$endgroup$
– C Monsour
May 6 '18 at 5:55
$begingroup$
Jacky.Very nice, clear and concise.Thanks!
$endgroup$
– Peter Szilas
May 6 '18 at 6:51
$begingroup$
Jacky.Very nice, clear and concise.Thanks!
$endgroup$
– Peter Szilas
May 6 '18 at 6:51
add a comment |
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$begingroup$
Fred, if this question is from a Calculus II course, my question is, how are you supposed to come up with an answer using the techniques available from Calculus II material? You may want to convey that to your professor. Which brings me to the second question. From which textbook is this exercise? Just curious
$endgroup$
– imranfat
May 6 '18 at 5:49
$begingroup$
@imranfat This is from a practice sheet made by the instructor. I'm his TA for the course. Doing this on the board, I didn't immediately know what to do; and thinking about it on and off, I'm convinced it's a mistake, as I don't see any immediate way to apply the traditional Calculus II material. Regardless, it's an interesting question that's been bugging me.
$endgroup$
– Fred
May 6 '18 at 5:52
$begingroup$
Well, you can ask your professor what test he expects the students doe carry out. He doesn't have to show the work; you can share that with us. I am just curious...
$endgroup$
– imranfat
May 6 '18 at 16:11