Studying the convergence of $sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$ The Next CEO of...
Strange use of "whether ... than ..." in official text
Does the Idaho Potato Commission associate potato skins with healthy eating?
Why do we say 'Un seul M' and not 'Une seule M' even though M is a "consonne"
Vector calculus integration identity problem
Help! I cannot understand this game’s notations!
Is there an equivalent of cd - for cp or mv
How did Beeri the Hittite come up with naming his daughter Yehudit?
Is there a way to save my career from absolute disaster?
What is the process for cleansing a very negative action
Won the lottery - how do I keep the money?
How to get the last not-null value in an ordered column of a huge table?
Can you teleport closer to a creature you are Frightened of?
Where do students learn to solve polynomial equations these days?
Yu-Gi-Oh cards in Python 3
The Ultimate Number Sequence Puzzle
Spaces in which all closed sets are regular closed
Purpose of level-shifter with same in and out voltages
what's the use of '% to gdp' type of variables?
In the "Harry Potter and the Order of the Phoenix" video game, what potion is used to sabotage Umbridge's speakers?
Does higher Oxidation/ reduction potential translate to higher energy storage in battery?
What are the unusually-enlarged wing sections on this P-38 Lightning?
Can I board the first leg of the flight without having final country's visa?
How to use ReplaceAll on an expression that contains a rule
Towers in the ocean; How deep can they be built?
Studying the convergence of $sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$
The Next CEO of Stack OverflowConvergence of $sumlimits_{n=2}^infty frac{1}{n^alpha ln^beta (n)} $ for nonnegative $alpha$ and $beta$Test for convergence $sum_{n=1}^{infty} frac{1}{2^sqrt{n}}$Studying the absolute and conditional convergence of $sum_{n=1}^{infty} frac{(-1)^n}{n^a(ln(n))^3}$Does the series $sum_limits{n=1}^{infty}frac{z^n}{a^{sqrt{n}}}$ converge for $a<1$?Convergence using Weierstrass's test $sum_limits{n=1}^{infty}frac{n!}{a^{n^2}}z^n$Study the convergence of $sum_limits{n=1}^{infty}frac{(-1)^n}{(x+n)^p}$Convergence domain of $sum_limits{n=1}^{infty}(-1)^{n+1}frac{1}{n^x}$?Absolute convergence of $sum_limits{n=1}^{infty}(-1)^{n+1}e^{-nsin x}$Find the convergence domain of the series $sum_limits{n=1}^{infty}frac{(-1)^{n-1}}{n3^n(x-5)^n}$Studying the convergence of $sum_limits{n=1}^{infty}frac{cos(in)}{2^n}$
$begingroup$
Study the convergence of the following series:$$sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$$
I thought of applying the root test:
$lim_{ntoinfty}({frac{e^{2in}}{nsqrt{n}}})^{frac{1}{n}}=lim_{ntoinfty}frac{e^{2i}}{n^{frac{3}{2n}}}=e^i$
However is $e^i>1$?
Questions:
1) Is the application of the root test right? What conclusion can I take?
2) Which would be other alternative ways of solving the question?
Thanks in advance!
sequences-and-series complex-analysis
asked Mar 17 at 17:05
Pedro GomesPedro Gomes
1,9822721
$endgroup$
add a comment |
$begingroup$
Study the convergence of the following series:$$sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$$
I thought of applying the root test:
$lim_{ntoinfty}({frac{e^{2in}}{nsqrt{n}}})^{frac{1}{n}}=lim_{ntoinfty}frac{e^{2i}}{n^{frac{3}{2n}}}=e^i$
However is $e^i>1$?
Questions:
1) Is the application of the root test right? What conclusion can I take?
2) Which would be other alternative ways of solving the question?
Thanks in advance!
sequences-and-series complex-analysis
asked Mar 17 at 17:05
Pedro GomesPedro Gomes
1,9822721
$endgroup$
$begingroup$
No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
$endgroup$
– Conrad
Mar 17 at 17:10
$begingroup$
Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
$endgroup$
– xbh
Mar 17 at 17:21
1
$begingroup$
@Conrad The root test DOES apply, but is inconclusive.
$endgroup$
– Mark Viola
Mar 17 at 17:44
$begingroup$
???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
$endgroup$
– Conrad
Mar 17 at 18:15
add a comment |
$begingroup$
Study the convergence of the following series:$$sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$$
I thought of applying the root test:
$lim_{ntoinfty}({frac{e^{2in}}{nsqrt{n}}})^{frac{1}{n}}=lim_{ntoinfty}frac{e^{2i}}{n^{frac{3}{2n}}}=e^i$
However is $e^i>1$?
Questions:
1) Is the application of the root test right? What conclusion can I take?
2) Which would be other alternative ways of solving the question?
Thanks in advance!
sequences-and-series complex-analysis
asked Mar 17 at 17:05
Pedro GomesPedro Gomes
1,9822721
$endgroup$
Study the convergence of the following series:$$sum_limits{n=1}^{infty}frac{e^{2in}}{nsqrt{n}}$$
I thought of applying the root test:
$lim_{ntoinfty}({frac{e^{2in}}{nsqrt{n}}})^{frac{1}{n}}=lim_{ntoinfty}frac{e^{2i}}{n^{frac{3}{2n}}}=e^i$
However is $e^i>1$?
Questions:
1) Is the application of the root test right? What conclusion can I take?
2) Which would be other alternative ways of solving the question?
Thanks in advance!
sequences-and-series complex-analysis
sequences-and-series complex-analysis
asked Mar 17 at 17:05
Pedro GomesPedro Gomes
1,9822721
asked Mar 17 at 17:05
Pedro GomesPedro Gomes
1,9822721
asked Mar 17 at 17:05
Pedro GomesPedro Gomes
1,9822721
asked Mar 17 at 17:05
Pedro GomesPedro Gomes
1,9822721
asked Mar 17 at 17:05
Pedro GomesPedro Gomes
1,9822721
1,9822721
$begingroup$
No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
$endgroup$
– Conrad
Mar 17 at 17:10
$begingroup$
Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
$endgroup$
– xbh
Mar 17 at 17:21
1
$begingroup$
@Conrad The root test DOES apply, but is inconclusive.
$endgroup$
– Mark Viola
Mar 17 at 17:44
$begingroup$
???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
$endgroup$
– Conrad
Mar 17 at 18:15
add a comment |
$begingroup$
No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
$endgroup$
– Conrad
Mar 17 at 17:10
$begingroup$
Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
$endgroup$
– xbh
Mar 17 at 17:21
1
$begingroup$
@Conrad The root test DOES apply, but is inconclusive.
$endgroup$
– Mark Viola
Mar 17 at 17:44
$begingroup$
???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
$endgroup$
– Conrad
Mar 17 at 18:15
$begingroup$
No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
$endgroup$
– Conrad
Mar 17 at 17:10
$begingroup$
No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
$endgroup$
– Conrad
Mar 17 at 17:10
$begingroup$
Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
$endgroup$
– xbh
Mar 17 at 17:21
$begingroup$
Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
$endgroup$
– xbh
Mar 17 at 17:21
1
1
$begingroup$
@Conrad The root test DOES apply, but is inconclusive.
$endgroup$
– Mark Viola
Mar 17 at 17:44
$begingroup$
@Conrad The root test DOES apply, but is inconclusive.
$endgroup$
– Mark Viola
Mar 17 at 17:44
$begingroup$
???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
$endgroup$
– Conrad
Mar 17 at 18:15
$begingroup$
???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
$endgroup$
– Conrad
Mar 17 at 18:15
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.
On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.
answered Mar 17 at 17:10
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
Application of the ROOT TEST shows
$$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$
So, the root test is inconclusive.
On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).
The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!
answered Mar 17 at 17:41
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
@pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 25 at 17:55
add a comment |
$begingroup$
As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.
Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.
EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.
answered Mar 17 at 17:46
Ryan GouldenRyan Goulden
495310
$endgroup$
add a comment |
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151776%2fstudying-the-convergence-of-sum-limitsn-1-infty-frace2inn-sqrtn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.
On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.
answered Mar 17 at 17:10
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.
On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.
answered Mar 17 at 17:10
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
add a comment |
$begingroup$
When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.
On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.
answered Mar 17 at 17:10
José Carlos SantosJosé Carlos Santos
171k23132240
$endgroup$
When you apply the root test to a series $displaystylesum_{n=1}^infty a_n$, what you should compute is the limit $displaystylelim_{ntoinfty}sqrt[n]{lvert a_nrvert}$. That is not what you did.
On the other hand, $(forall ninmathbb N):leftlvertdfrac{e^{2in}}{nsqrt n}rightrvert=dfrac1{n^{3/2}}$. Since the series $displaystylesum_{n=1}^inftydfrac1{n^{3/2}}$ converges (by the integral test), your series converges absolutely. In particular, it converges.
answered Mar 17 at 17:10
José Carlos SantosJosé Carlos Santos
171k23132240
edited Mar 17 at 17:38
answered Mar 17 at 17:10
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 17 at 17:10
José Carlos SantosJosé Carlos Santos
171k23132240
answered Mar 17 at 17:10
José Carlos SantosJosé Carlos Santos
171k23132240
171k23132240
add a comment |
add a comment |
$begingroup$
Application of the ROOT TEST shows
$$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$
So, the root test is inconclusive.
On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).
The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!
answered Mar 17 at 17:41
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
@pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 25 at 17:55
add a comment |
$begingroup$
Application of the ROOT TEST shows
$$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$
So, the root test is inconclusive.
On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).
The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!
answered Mar 17 at 17:41
Mark ViolaMark Viola
134k1278176
$endgroup$
$begingroup$
@pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 25 at 17:55
add a comment |
$begingroup$
Application of the ROOT TEST shows
$$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$
So, the root test is inconclusive.
On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).
The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!
answered Mar 17 at 17:41
Mark ViolaMark Viola
134k1278176
$endgroup$
Application of the ROOT TEST shows
$$limsup_{ntoinfty}sqrt[n]{left|frac{e^{i2n}}{n^{3/2}}right|}=lim_{nto infty}sqrt[n]{n^{-3/2}}=1$$
So, the root test is inconclusive.
On the other hand, Dirichlet's Test is applicable since $frac{1}{n^{3/2}}$ monotonically decreases to $0$ and for all $N$, $left|sum_{n=1}^N e^{i2n} right|$ is bounded (in fact it is bounded by $csc(1)$).
The power of Dirichlet's test is not of full display here since $sum_{n=1}^infty frac{e^{i2n}}{n^{3/2}}$ converges absolutely. But note that Dirichlet's test guarantees that the series $sum_{n=1}^infty frac{e^{i2n}}{n^a}$ converges for all $a>0$!
answered Mar 17 at 17:41
Mark ViolaMark Viola
134k1278176
answered Mar 17 at 17:41
Mark ViolaMark Viola
134k1278176
answered Mar 17 at 17:41
Mark ViolaMark Viola
134k1278176
answered Mar 17 at 17:41
Mark ViolaMark Viola
134k1278176
134k1278176
$begingroup$
@pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 25 at 17:55
add a comment |
$begingroup$
@pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 25 at 17:55
$begingroup$
@pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 25 at 17:55
$begingroup$
@pedrogomes Please let me know how I can improve my answer. I really want to give you the best answer I can.
$endgroup$
– Mark Viola
Mar 25 at 17:55
add a comment |
$begingroup$
As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.
Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.
EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.
answered Mar 17 at 17:46
Ryan GouldenRyan Goulden
495310
$endgroup$
add a comment |
$begingroup$
As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.
Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.
EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.
answered Mar 17 at 17:46
Ryan GouldenRyan Goulden
495310
$endgroup$
add a comment |
$begingroup$
As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.
Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.
EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.
answered Mar 17 at 17:46
Ryan GouldenRyan Goulden
495310
$endgroup$
As an alternative, if you know that $sum frac{1}{n^c}$ converges for $c>1$, and that $e^{2in} = cos(2n) + isin(2n)$ then this isn't too hard.
Consider the series $$sum |frac{cos(2n)}{n√n}| < sum frac{1}{n^{frac{3}{2}}},$$ and from here, we may conclude the original series converges.
EDIT: additionally, to answer your question about $e^{i} > 1$: this isn't true, but it contrarily, is not true that $e^{i}<1$. $e^{i} = cos(1) + isin(1)$, and you are trying to compare this to a strictly real number. It is tantamount to asking if $i$ is greater than $1$. There's no total ordering on the complex numbers, so these kinds of comparisons don't make sense. Instead, you could consider $|e^{i}|$ which is, in fact, equal to $1$.
answered Mar 17 at 17:46
Ryan GouldenRyan Goulden
495310
edited Mar 17 at 17:58
answered Mar 17 at 17:46
Ryan GouldenRyan Goulden
495310
answered Mar 17 at 17:46
Ryan GouldenRyan Goulden
495310
answered Mar 17 at 17:46
Ryan GouldenRyan Goulden
495310
495310
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3151776%2fstudying-the-convergence-of-sum-limitsn-1-infty-frace2inn-sqrtn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
No! Root test doesn't apply. The way to do it is taking absolute values and noting that $|e^{2in}|=1$
$endgroup$
– Conrad
Mar 17 at 17:10
$begingroup$
Also that is not the correct way to use the root test. The quantity for used in testing is $vert a_n vert ^{1/n}$, so actually the limit is just $1$, since $vert mathrm e ^{mathrm i 2n} vert = 1$.
$endgroup$
– xbh
Mar 17 at 17:21
1
$begingroup$
@Conrad The root test DOES apply, but is inconclusive.
$endgroup$
– Mark Viola
Mar 17 at 17:44
$begingroup$
???? If it's inconclusive how does it apply? Maybe it's a matter of semantics but if a method doesn't solve a problem, it doesn't really apply the way I see it
$endgroup$
– Conrad
Mar 17 at 18:15