Comparing to a constant with the limit comparison test The Next CEO of Stack Overflowlimit of...
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Comparing to a constant with the limit comparison test
The Next CEO of Stack Overflowlimit of a sequence; using comparison testTesting Convergence With Limit Comparison TestCalculus II: Comparison Test for DivergenceShould I use the comparison test for the following series?The Comparison Test for Seriescomparison or limit comparison test..?Incorrect comparison test on seriesComparison test $sum_{n=1}^{infty} frac{1}{2n+7}$Comparison test of series with ln functionWhy do I have to use L'Hopital in limit comparison test for $sum_{n=1}^{infty} sinleft(frac{1}{n}right)$
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Recently in my calculus class, we covered the limit comparison test for infinite series (among other things related to sequences and series). We have also covered improper integrals earlier this semester. One of the homework questions asks us to "calculate" $$sum_{n=1}^infty a_n::text{where}::a_n=arctanleft(frac {n^2}{n^2+4}right).$$
I noted that, as $a_nto infty$, $arctanleft(frac {n^2}{n^2+4}right)approx arctan(1)$, and used that to compare using the limit comparison test, resulting in $$lim_limits{ntoinfty}frac{arctanleft(frac {n^2}{n^2+4}right)}{arctan(1)} = 1,$$ so I concluded that $a_n$ diverges, since $$sum_{n=1}^infty arctan(1) = infty.$$
I was wondering if, both in this specific case and in general, using a constant with the limit comparison test was even allowed, since up until now we have been comparing to similar $f(x)$ or $b_n$ for integrals and series respectively.
calculus sequences-and-series improper-integrals
edited Mar 18 at 2:47
Robert Howard
2,2933935
asked Mar 18 at 2:36
TropingenieTropingenie
105
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add a comment |
$begingroup$
Recently in my calculus class, we covered the limit comparison test for infinite series (among other things related to sequences and series). We have also covered improper integrals earlier this semester. One of the homework questions asks us to "calculate" $$sum_{n=1}^infty a_n::text{where}::a_n=arctanleft(frac {n^2}{n^2+4}right).$$
I noted that, as $a_nto infty$, $arctanleft(frac {n^2}{n^2+4}right)approx arctan(1)$, and used that to compare using the limit comparison test, resulting in $$lim_limits{ntoinfty}frac{arctanleft(frac {n^2}{n^2+4}right)}{arctan(1)} = 1,$$ so I concluded that $a_n$ diverges, since $$sum_{n=1}^infty arctan(1) = infty.$$
I was wondering if, both in this specific case and in general, using a constant with the limit comparison test was even allowed, since up until now we have been comparing to similar $f(x)$ or $b_n$ for integrals and series respectively.
calculus sequences-and-series improper-integrals
edited Mar 18 at 2:47
Robert Howard
2,2933935
asked Mar 18 at 2:36
TropingenieTropingenie
105
$endgroup$
add a comment |
$begingroup$
Recently in my calculus class, we covered the limit comparison test for infinite series (among other things related to sequences and series). We have also covered improper integrals earlier this semester. One of the homework questions asks us to "calculate" $$sum_{n=1}^infty a_n::text{where}::a_n=arctanleft(frac {n^2}{n^2+4}right).$$
I noted that, as $a_nto infty$, $arctanleft(frac {n^2}{n^2+4}right)approx arctan(1)$, and used that to compare using the limit comparison test, resulting in $$lim_limits{ntoinfty}frac{arctanleft(frac {n^2}{n^2+4}right)}{arctan(1)} = 1,$$ so I concluded that $a_n$ diverges, since $$sum_{n=1}^infty arctan(1) = infty.$$
I was wondering if, both in this specific case and in general, using a constant with the limit comparison test was even allowed, since up until now we have been comparing to similar $f(x)$ or $b_n$ for integrals and series respectively.
calculus sequences-and-series improper-integrals
edited Mar 18 at 2:47
Robert Howard
2,2933935
asked Mar 18 at 2:36
TropingenieTropingenie
105
$endgroup$
Recently in my calculus class, we covered the limit comparison test for infinite series (among other things related to sequences and series). We have also covered improper integrals earlier this semester. One of the homework questions asks us to "calculate" $$sum_{n=1}^infty a_n::text{where}::a_n=arctanleft(frac {n^2}{n^2+4}right).$$
I noted that, as $a_nto infty$, $arctanleft(frac {n^2}{n^2+4}right)approx arctan(1)$, and used that to compare using the limit comparison test, resulting in $$lim_limits{ntoinfty}frac{arctanleft(frac {n^2}{n^2+4}right)}{arctan(1)} = 1,$$ so I concluded that $a_n$ diverges, since $$sum_{n=1}^infty arctan(1) = infty.$$
I was wondering if, both in this specific case and in general, using a constant with the limit comparison test was even allowed, since up until now we have been comparing to similar $f(x)$ or $b_n$ for integrals and series respectively.
calculus sequences-and-series improper-integrals
calculus sequences-and-series improper-integrals
edited Mar 18 at 2:47
Robert Howard
2,2933935
asked Mar 18 at 2:36
TropingenieTropingenie
105
edited Mar 18 at 2:47
Robert Howard
2,2933935
asked Mar 18 at 2:36
TropingenieTropingenie
105
edited Mar 18 at 2:47
Robert Howard
2,2933935
edited Mar 18 at 2:47
Robert Howard
2,2933935
edited Mar 18 at 2:47
Robert Howard
2,2933935
2,2933935
asked Mar 18 at 2:36
TropingenieTropingenie
105
asked Mar 18 at 2:36
TropingenieTropingenie
105
asked Mar 18 at 2:36
TropingenieTropingenie
105
105
add a comment |
add a comment |
1 Answer
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Yes: the function $f(x) = arctan 1$ (independent of $x$) is a perfectly valid function.
Note that in such cases, the Test for Divergence will work more quickly, since the summand does not tend to $0$.
answered Mar 18 at 3:27
Greg MartinGreg Martin
36.5k23565
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add a comment |
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$begingroup$
Yes: the function $f(x) = arctan 1$ (independent of $x$) is a perfectly valid function.
Note that in such cases, the Test for Divergence will work more quickly, since the summand does not tend to $0$.
answered Mar 18 at 3:27
Greg MartinGreg Martin
36.5k23565
$endgroup$
add a comment |
$begingroup$
Yes: the function $f(x) = arctan 1$ (independent of $x$) is a perfectly valid function.
Note that in such cases, the Test for Divergence will work more quickly, since the summand does not tend to $0$.
answered Mar 18 at 3:27
Greg MartinGreg Martin
36.5k23565
$endgroup$
add a comment |
$begingroup$
Yes: the function $f(x) = arctan 1$ (independent of $x$) is a perfectly valid function.
Note that in such cases, the Test for Divergence will work more quickly, since the summand does not tend to $0$.
answered Mar 18 at 3:27
Greg MartinGreg Martin
36.5k23565
$endgroup$
Yes: the function $f(x) = arctan 1$ (independent of $x$) is a perfectly valid function.
Note that in such cases, the Test for Divergence will work more quickly, since the summand does not tend to $0$.
answered Mar 18 at 3:27
Greg MartinGreg Martin
36.5k23565
answered Mar 18 at 3:27
Greg MartinGreg Martin
36.5k23565
answered Mar 18 at 3:27
Greg MartinGreg Martin
36.5k23565
answered Mar 18 at 3:27
Greg MartinGreg Martin
36.5k23565
36.5k23565
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