Find the value of infinite series $sum_{n=1}^{infty} tan^{-1}(2/n^2)$
5
$begingroup$
Find the value of infinite series
$$ sum_{n=1}^{infty} tan^{-1}left(frac{2}{n^2} right) $$
I tried to find sequence of partial sums and tried to find the limit of that sequence. But I didn't get the answer.
real-analysis sequences-and-series functions limits-without-lhopital
edited Sep 13 '17 at 7:14
Sangchul Lee
96.2k12171281
asked Sep 13 '17 at 7:09
AnanthakrishnanAnanthakrishnan
847
$endgroup$
add a comment |
5
$begingroup$
Find the value of infinite series
$$ sum_{n=1}^{infty} tan^{-1}left(frac{2}{n^2} right) $$
I tried to find sequence of partial sums and tried to find the limit of that sequence. But I didn't get the answer.
real-analysis sequences-and-series functions limits-without-lhopital
edited Sep 13 '17 at 7:14
Sangchul Lee
96.2k12171281
asked Sep 13 '17 at 7:09
AnanthakrishnanAnanthakrishnan
847
$endgroup$
3
$begingroup$
If I remember correctly, this problem goes back to Ramanujan. And it can be computed using telescoping technique combined with the following identity: $$ tan^{-1}left(frac{1}{n-1}right) - tan^{-1}left(frac{1}{n+1}right) = arctanleft(frac{2}{n^2}right). $$
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:17
1
$begingroup$
Related : math.stackexchange.com/questions/144944/…, math.stackexchange.com/questions/415512/… math.stackexchange.com/questions/617032/…
$endgroup$
– lab bhattacharjee
Sep 13 '17 at 7:57
$begingroup$
Also an elementary approach is given here. It is works from the vein that @SangchulLee pointed out.
$endgroup$
– Mason
Dec 27 '18 at 15:50
add a comment |
5
5
5
2
$begingroup$
Find the value of infinite series
$$ sum_{n=1}^{infty} tan^{-1}left(frac{2}{n^2} right) $$
I tried to find sequence of partial sums and tried to find the limit of that sequence. But I didn't get the answer.
real-analysis sequences-and-series functions limits-without-lhopital
edited Sep 13 '17 at 7:14
Sangchul Lee
96.2k12171281
asked Sep 13 '17 at 7:09
AnanthakrishnanAnanthakrishnan
847
$endgroup$
Find the value of infinite series
$$ sum_{n=1}^{infty} tan^{-1}left(frac{2}{n^2} right) $$
I tried to find sequence of partial sums and tried to find the limit of that sequence. But I didn't get the answer.
real-analysis sequences-and-series functions limits-without-lhopital
real-analysis sequences-and-series functions limits-without-lhopital
edited Sep 13 '17 at 7:14
Sangchul Lee
96.2k12171281
asked Sep 13 '17 at 7:09
AnanthakrishnanAnanthakrishnan
847
edited Sep 13 '17 at 7:14
Sangchul Lee
96.2k12171281
asked Sep 13 '17 at 7:09
AnanthakrishnanAnanthakrishnan
847
edited Sep 13 '17 at 7:14
Sangchul Lee
96.2k12171281
edited Sep 13 '17 at 7:14
Sangchul Lee
96.2k12171281
edited Sep 13 '17 at 7:14
Sangchul Lee
96.2k12171281
96.2k12171281
asked Sep 13 '17 at 7:09
AnanthakrishnanAnanthakrishnan
847
asked Sep 13 '17 at 7:09
AnanthakrishnanAnanthakrishnan
847
asked Sep 13 '17 at 7:09
AnanthakrishnanAnanthakrishnan
847
847
3
$begingroup$
If I remember correctly, this problem goes back to Ramanujan. And it can be computed using telescoping technique combined with the following identity: $$ tan^{-1}left(frac{1}{n-1}right) - tan^{-1}left(frac{1}{n+1}right) = arctanleft(frac{2}{n^2}right). $$
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:17
1
$begingroup$
Related : math.stackexchange.com/questions/144944/…, math.stackexchange.com/questions/415512/… math.stackexchange.com/questions/617032/…
$endgroup$
– lab bhattacharjee
Sep 13 '17 at 7:57
$begingroup$
Also an elementary approach is given here. It is works from the vein that @SangchulLee pointed out.
$endgroup$
– Mason
Dec 27 '18 at 15:50
add a comment |
3
$begingroup$
If I remember correctly, this problem goes back to Ramanujan. And it can be computed using telescoping technique combined with the following identity: $$ tan^{-1}left(frac{1}{n-1}right) - tan^{-1}left(frac{1}{n+1}right) = arctanleft(frac{2}{n^2}right). $$
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:17
1
$begingroup$
Related : math.stackexchange.com/questions/144944/…, math.stackexchange.com/questions/415512/… math.stackexchange.com/questions/617032/…
$endgroup$
– lab bhattacharjee
Sep 13 '17 at 7:57
$begingroup$
Also an elementary approach is given here. It is works from the vein that @SangchulLee pointed out.
$endgroup$
– Mason
Dec 27 '18 at 15:50
3
3
$begingroup$
If I remember correctly, this problem goes back to Ramanujan. And it can be computed using telescoping technique combined with the following identity: $$ tan^{-1}left(frac{1}{n-1}right) - tan^{-1}left(frac{1}{n+1}right) = arctanleft(frac{2}{n^2}right). $$
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:17
$begingroup$
If I remember correctly, this problem goes back to Ramanujan. And it can be computed using telescoping technique combined with the following identity: $$ tan^{-1}left(frac{1}{n-1}right) - tan^{-1}left(frac{1}{n+1}right) = arctanleft(frac{2}{n^2}right). $$
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:17
1
1
$begingroup$
Related : math.stackexchange.com/questions/144944/…, math.stackexchange.com/questions/415512/… math.stackexchange.com/questions/617032/…
$endgroup$
– lab bhattacharjee
Sep 13 '17 at 7:57
$begingroup$
Related : math.stackexchange.com/questions/144944/…, math.stackexchange.com/questions/415512/… math.stackexchange.com/questions/617032/…
$endgroup$
– lab bhattacharjee
Sep 13 '17 at 7:57
$begingroup$
Also an elementary approach is given here. It is works from the vein that @SangchulLee pointed out.
$endgroup$
– Mason
Dec 27 '18 at 15:50
$begingroup$
Also an elementary approach is given here. It is works from the vein that @SangchulLee pointed out.
$endgroup$
– Mason
Dec 27 '18 at 15:50
add a comment |
1 Answer
1
active
oldest
votes
7
$begingroup$
It's the argument of the complex number
$$prod_{n=1}^inftyleft(1+frac{2i}{n^2}right)
=prod_{n=1}^inftyfrac{(n-1+i)(n+1-i)}{n^2}
=prod_{n=1}^inftyfrac{(n-1+i)(n^2+2n)}{n^2(n+1+i)}.$$
This infinite product telescopes.
answered Sep 13 '17 at 7:21
Lord Shark the UnknownLord Shark the Unknown
107k1162134
$endgroup$
$begingroup$
Briefly, how did you get the first step? :(
$endgroup$
– Joao Noch
Sep 13 '17 at 7:30
$begingroup$
@JoaoNoch, It follows from $ arctan(x) = arg(1+ix)$. Perhaps the only non-trivial part is that the argument of the product above determines OP's sum only modulo $2pi$. But an estimate $$sum_{n=1}^{infty} arctan(2/n^2) leq frac{pi}{2} + 2(zeta(2)-1) < pi$$ shows that the principal argument function is enough for our purpose.
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:50
add a comment |
Your Answer
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
});
}
});
draft saved
draft discarded
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%2f2427602%2ffind-the-value-of-infinite-series-sum-n-1-infty-tan-12-n2%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
7
$begingroup$
It's the argument of the complex number
$$prod_{n=1}^inftyleft(1+frac{2i}{n^2}right)
=prod_{n=1}^inftyfrac{(n-1+i)(n+1-i)}{n^2}
=prod_{n=1}^inftyfrac{(n-1+i)(n^2+2n)}{n^2(n+1+i)}.$$
This infinite product telescopes.
answered Sep 13 '17 at 7:21
Lord Shark the UnknownLord Shark the Unknown
107k1162134
$endgroup$
$begingroup$
Briefly, how did you get the first step? :(
$endgroup$
– Joao Noch
Sep 13 '17 at 7:30
$begingroup$
@JoaoNoch, It follows from $ arctan(x) = arg(1+ix)$. Perhaps the only non-trivial part is that the argument of the product above determines OP's sum only modulo $2pi$. But an estimate $$sum_{n=1}^{infty} arctan(2/n^2) leq frac{pi}{2} + 2(zeta(2)-1) < pi$$ shows that the principal argument function is enough for our purpose.
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:50
add a comment |
7
$begingroup$
It's the argument of the complex number
$$prod_{n=1}^inftyleft(1+frac{2i}{n^2}right)
=prod_{n=1}^inftyfrac{(n-1+i)(n+1-i)}{n^2}
=prod_{n=1}^inftyfrac{(n-1+i)(n^2+2n)}{n^2(n+1+i)}.$$
This infinite product telescopes.
answered Sep 13 '17 at 7:21
Lord Shark the UnknownLord Shark the Unknown
107k1162134
$endgroup$
$begingroup$
Briefly, how did you get the first step? :(
$endgroup$
– Joao Noch
Sep 13 '17 at 7:30
$begingroup$
@JoaoNoch, It follows from $ arctan(x) = arg(1+ix)$. Perhaps the only non-trivial part is that the argument of the product above determines OP's sum only modulo $2pi$. But an estimate $$sum_{n=1}^{infty} arctan(2/n^2) leq frac{pi}{2} + 2(zeta(2)-1) < pi$$ shows that the principal argument function is enough for our purpose.
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:50
add a comment |
7
7
7
$begingroup$
It's the argument of the complex number
$$prod_{n=1}^inftyleft(1+frac{2i}{n^2}right)
=prod_{n=1}^inftyfrac{(n-1+i)(n+1-i)}{n^2}
=prod_{n=1}^inftyfrac{(n-1+i)(n^2+2n)}{n^2(n+1+i)}.$$
This infinite product telescopes.
answered Sep 13 '17 at 7:21
Lord Shark the UnknownLord Shark the Unknown
107k1162134
$endgroup$
It's the argument of the complex number
$$prod_{n=1}^inftyleft(1+frac{2i}{n^2}right)
=prod_{n=1}^inftyfrac{(n-1+i)(n+1-i)}{n^2}
=prod_{n=1}^inftyfrac{(n-1+i)(n^2+2n)}{n^2(n+1+i)}.$$
This infinite product telescopes.
answered Sep 13 '17 at 7:21
Lord Shark the UnknownLord Shark the Unknown
107k1162134
answered Sep 13 '17 at 7:21
Lord Shark the UnknownLord Shark the Unknown
107k1162134
answered Sep 13 '17 at 7:21
Lord Shark the UnknownLord Shark the Unknown
107k1162134
answered Sep 13 '17 at 7:21
Lord Shark the UnknownLord Shark the Unknown
107k1162134
107k1162134
$begingroup$
Briefly, how did you get the first step? :(
$endgroup$
– Joao Noch
Sep 13 '17 at 7:30
$begingroup$
@JoaoNoch, It follows from $ arctan(x) = arg(1+ix)$. Perhaps the only non-trivial part is that the argument of the product above determines OP's sum only modulo $2pi$. But an estimate $$sum_{n=1}^{infty} arctan(2/n^2) leq frac{pi}{2} + 2(zeta(2)-1) < pi$$ shows that the principal argument function is enough for our purpose.
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:50
add a comment |
$begingroup$
Briefly, how did you get the first step? :(
$endgroup$
– Joao Noch
Sep 13 '17 at 7:30
$begingroup$
@JoaoNoch, It follows from $ arctan(x) = arg(1+ix)$. Perhaps the only non-trivial part is that the argument of the product above determines OP's sum only modulo $2pi$. But an estimate $$sum_{n=1}^{infty} arctan(2/n^2) leq frac{pi}{2} + 2(zeta(2)-1) < pi$$ shows that the principal argument function is enough for our purpose.
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:50
$begingroup$
Briefly, how did you get the first step? :(
$endgroup$
– Joao Noch
Sep 13 '17 at 7:30
$begingroup$
Briefly, how did you get the first step? :(
$endgroup$
– Joao Noch
Sep 13 '17 at 7:30
$begingroup$
@JoaoNoch, It follows from $ arctan(x) = arg(1+ix)$. Perhaps the only non-trivial part is that the argument of the product above determines OP's sum only modulo $2pi$. But an estimate $$sum_{n=1}^{infty} arctan(2/n^2) leq frac{pi}{2} + 2(zeta(2)-1) < pi$$ shows that the principal argument function is enough for our purpose.
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:50
$begingroup$
@JoaoNoch, It follows from $ arctan(x) = arg(1+ix)$. Perhaps the only non-trivial part is that the argument of the product above determines OP's sum only modulo $2pi$. But an estimate $$sum_{n=1}^{infty} arctan(2/n^2) leq frac{pi}{2} + 2(zeta(2)-1) < pi$$ shows that the principal argument function is enough for our purpose.
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:50
add a comment |
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f2427602%2ffind-the-value-of-infinite-series-sum-n-1-infty-tan-12-n2%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
3
$begingroup$
If I remember correctly, this problem goes back to Ramanujan. And it can be computed using telescoping technique combined with the following identity: $$ tan^{-1}left(frac{1}{n-1}right) - tan^{-1}left(frac{1}{n+1}right) = arctanleft(frac{2}{n^2}right). $$
$endgroup$
– Sangchul Lee
Sep 13 '17 at 7:17
1
$begingroup$
Related : math.stackexchange.com/questions/144944/…, math.stackexchange.com/questions/415512/… math.stackexchange.com/questions/617032/…
$endgroup$
– lab bhattacharjee
Sep 13 '17 at 7:57
$begingroup$
Also an elementary approach is given here. It is works from the vein that @SangchulLee pointed out.
$endgroup$
– Mason
Dec 27 '18 at 15:50