Evaluate $sum_{n=1}^{infty}arctan(frac{2}{n^2})$?

2

$begingroup$

$$begin{align}
sum_{n=1}^{infty}arctanleft(frac{2}{n^2}right)&=sum_{n=1}^{infty}arctanleft(frac{(n+1)-(n-1)}{1+(n+1)(n-1)}right)\
&=sum_{n=1}^{infty}arctan(n+1)-arctan(n-1)\
&= fracpi4
end{align}$$

Is this correct?

sequences-and-series

share|cite|improve this question

edited Jan 31 at 2:28

Larry

2,53031131

asked Jan 31 at 2:04

AbhayAbhay

3789

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the process is good and correct.
  $endgroup$
  – Offlaw
  Jan 31 at 2:07
  

  
  
  
  

add a comment | 

2

$begingroup$

$$begin{align}
sum_{n=1}^{infty}arctanleft(frac{2}{n^2}right)&=sum_{n=1}^{infty}arctanleft(frac{(n+1)-(n-1)}{1+(n+1)(n-1)}right)\
&=sum_{n=1}^{infty}arctan(n+1)-arctan(n-1)\
&= fracpi4
end{align}$$

Is this correct?

sequences-and-series

share|cite|improve this question

edited Jan 31 at 2:28

Larry

2,53031131

asked Jan 31 at 2:04

AbhayAbhay

3789

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the process is good and correct.
  $endgroup$
  – Offlaw
  Jan 31 at 2:07
  

  
  
  
  

add a comment | 

2

2

2

1

$begingroup$

$$begin{align}
sum_{n=1}^{infty}arctanleft(frac{2}{n^2}right)&=sum_{n=1}^{infty}arctanleft(frac{(n+1)-(n-1)}{1+(n+1)(n-1)}right)\
&=sum_{n=1}^{infty}arctan(n+1)-arctan(n-1)\
&= fracpi4
end{align}$$

Is this correct?

sequences-and-series

share|cite|improve this question

edited Jan 31 at 2:28

Larry

2,53031131

asked Jan 31 at 2:04

AbhayAbhay

3789

$endgroup$

$$begin{align}
sum_{n=1}^{infty}arctanleft(frac{2}{n^2}right)&=sum_{n=1}^{infty}arctanleft(frac{(n+1)-(n-1)}{1+(n+1)(n-1)}right)\
&=sum_{n=1}^{infty}arctan(n+1)-arctan(n-1)\
&= fracpi4
end{align}$$

Is this correct?

sequences-and-series

sequences-and-series

share|cite|improve this question

edited Jan 31 at 2:28

Larry

2,53031131

asked Jan 31 at 2:04

AbhayAbhay

3789

share|cite|improve this question

edited Jan 31 at 2:28

Larry

2,53031131

asked Jan 31 at 2:04

AbhayAbhay

3789

share|cite|improve this question

share|cite|improve this question

edited Jan 31 at 2:28

Larry

2,53031131

edited Jan 31 at 2:28

Larry

2,53031131

edited Jan 31 at 2:28

Larry

2,53031131

2,53031131

asked Jan 31 at 2:04

AbhayAbhay

3789

asked Jan 31 at 2:04

AbhayAbhay

3789

asked Jan 31 at 2:04

AbhayAbhay

3789

3789

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the process is good and correct.
  $endgroup$
  – Offlaw
  Jan 31 at 2:07
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I think the process is good and correct.
  $endgroup$
  – Offlaw
  Jan 31 at 2:07
  

  
  
  
  

$begingroup$
I think the process is good and correct.
$endgroup$
– Offlaw
Jan 31 at 2:07

$begingroup$
I think the process is good and correct.
$endgroup$
– Offlaw
Jan 31 at 2:07

add a comment | 

2 Answers
2

active

oldest

votes

2

$begingroup$

The series telescopes. The partial sum $S_N$ is equal to
begin{align}
S_N &= [arctan (2) - arctan (0)] + [arctan (3) - arctan (1)] + cdots\
& qquad cdots + [arctan (N) - arctan (N - 2)] + [arctan(N + 1) - arctan (N - 1)]\
&= - arctan (0) - arctan (1) + arctan (N) + arctan (N + 1)\
&= -frac{pi}{4} + arctan (N) + arctan (N + 1)
end{align}
So for the sum $S$ we have
begin{align}
S &= lim_{N to infty} S_N\
&= lim_{N to infty} left [- frac{pi}{4} + arctan(N) + arctan(N + 1) right ]\
&= -frac{pi}{4} + frac{pi}{2} + frac{pi}{2}\
&= frac{3 pi}{4}.
end{align}

share|cite|improve this answer

answered Jan 31 at 2:14

omegadotomegadot

6,2692829

$endgroup$

add a comment | 

0

$begingroup$

$$sum^{infty}_{n=1}bigg(tan^{-1}(n+1)-tan^{-1}(n)bigg)$$
$$+sum^{infty}_{n=1}bigg(tan^{-1}(n)-tan^{-1}(n-1)bigg)$$

Convert into Telescopic Series

$$=tan^{-1}(infty)-tan^{-1}(1)+tan^{-1}(infty)-tan^{-1}(0)=$$

$$=pi-frac{pi}{4}=frac{3pi}{4}.$$

share|cite|improve this answer

answered Jan 31 at 2:21

DXTDXT

5,8892732

$endgroup$

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
});


}
});

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094394%2fevaluate-sum-n-1-infty-arctan-frac2n2%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

2

$begingroup$

The series telescopes. The partial sum $S_N$ is equal to
begin{align}
S_N &= [arctan (2) - arctan (0)] + [arctan (3) - arctan (1)] + cdots\
& qquad cdots + [arctan (N) - arctan (N - 2)] + [arctan(N + 1) - arctan (N - 1)]\
&= - arctan (0) - arctan (1) + arctan (N) + arctan (N + 1)\
&= -frac{pi}{4} + arctan (N) + arctan (N + 1)
end{align}
So for the sum $S$ we have
begin{align}
S &= lim_{N to infty} S_N\
&= lim_{N to infty} left [- frac{pi}{4} + arctan(N) + arctan(N + 1) right ]\
&= -frac{pi}{4} + frac{pi}{2} + frac{pi}{2}\
&= frac{3 pi}{4}.
end{align}

share|cite|improve this answer

answered Jan 31 at 2:14

omegadotomegadot

6,2692829

$endgroup$

add a comment | 

2

$begingroup$

The series telescopes. The partial sum $S_N$ is equal to
begin{align}
S_N &= [arctan (2) - arctan (0)] + [arctan (3) - arctan (1)] + cdots\
& qquad cdots + [arctan (N) - arctan (N - 2)] + [arctan(N + 1) - arctan (N - 1)]\
&= - arctan (0) - arctan (1) + arctan (N) + arctan (N + 1)\
&= -frac{pi}{4} + arctan (N) + arctan (N + 1)
end{align}
So for the sum $S$ we have
begin{align}
S &= lim_{N to infty} S_N\
&= lim_{N to infty} left [- frac{pi}{4} + arctan(N) + arctan(N + 1) right ]\
&= -frac{pi}{4} + frac{pi}{2} + frac{pi}{2}\
&= frac{3 pi}{4}.
end{align}

share|cite|improve this answer

answered Jan 31 at 2:14

omegadotomegadot

6,2692829

$endgroup$

add a comment | 

2

2

2

$begingroup$

The series telescopes. The partial sum $S_N$ is equal to
begin{align}
S_N &= [arctan (2) - arctan (0)] + [arctan (3) - arctan (1)] + cdots\
& qquad cdots + [arctan (N) - arctan (N - 2)] + [arctan(N + 1) - arctan (N - 1)]\
&= - arctan (0) - arctan (1) + arctan (N) + arctan (N + 1)\
&= -frac{pi}{4} + arctan (N) + arctan (N + 1)
end{align}
So for the sum $S$ we have
begin{align}
S &= lim_{N to infty} S_N\
&= lim_{N to infty} left [- frac{pi}{4} + arctan(N) + arctan(N + 1) right ]\
&= -frac{pi}{4} + frac{pi}{2} + frac{pi}{2}\
&= frac{3 pi}{4}.
end{align}

share|cite|improve this answer

answered Jan 31 at 2:14

omegadotomegadot

6,2692829

$endgroup$

The series telescopes. The partial sum $S_N$ is equal to
begin{align}
S_N &= [arctan (2) - arctan (0)] + [arctan (3) - arctan (1)] + cdots\
& qquad cdots + [arctan (N) - arctan (N - 2)] + [arctan(N + 1) - arctan (N - 1)]\
&= - arctan (0) - arctan (1) + arctan (N) + arctan (N + 1)\
&= -frac{pi}{4} + arctan (N) + arctan (N + 1)
end{align}
So for the sum $S$ we have
begin{align}
S &= lim_{N to infty} S_N\
&= lim_{N to infty} left [- frac{pi}{4} + arctan(N) + arctan(N + 1) right ]\
&= -frac{pi}{4} + frac{pi}{2} + frac{pi}{2}\
&= frac{3 pi}{4}.
end{align}

share|cite|improve this answer

answered Jan 31 at 2:14

omegadotomegadot

6,2692829

share|cite|improve this answer

share|cite|improve this answer

answered Jan 31 at 2:14

omegadotomegadot

6,2692829

answered Jan 31 at 2:14

omegadotomegadot

6,2692829

answered Jan 31 at 2:14

omegadotomegadot

6,2692829

6,2692829

add a comment | 

add a comment | 

0

$begingroup$

$$sum^{infty}_{n=1}bigg(tan^{-1}(n+1)-tan^{-1}(n)bigg)$$
$$+sum^{infty}_{n=1}bigg(tan^{-1}(n)-tan^{-1}(n-1)bigg)$$

Convert into Telescopic Series

$$=tan^{-1}(infty)-tan^{-1}(1)+tan^{-1}(infty)-tan^{-1}(0)=$$

$$=pi-frac{pi}{4}=frac{3pi}{4}.$$

share|cite|improve this answer

answered Jan 31 at 2:21

DXTDXT

5,8892732

$endgroup$

add a comment | 

0

$begingroup$

$$sum^{infty}_{n=1}bigg(tan^{-1}(n+1)-tan^{-1}(n)bigg)$$
$$+sum^{infty}_{n=1}bigg(tan^{-1}(n)-tan^{-1}(n-1)bigg)$$

Convert into Telescopic Series

$$=tan^{-1}(infty)-tan^{-1}(1)+tan^{-1}(infty)-tan^{-1}(0)=$$

$$=pi-frac{pi}{4}=frac{3pi}{4}.$$

share|cite|improve this answer

answered Jan 31 at 2:21

DXTDXT

5,8892732

$endgroup$

add a comment | 

0

0

0

$begingroup$

$$sum^{infty}_{n=1}bigg(tan^{-1}(n+1)-tan^{-1}(n)bigg)$$
$$+sum^{infty}_{n=1}bigg(tan^{-1}(n)-tan^{-1}(n-1)bigg)$$

Convert into Telescopic Series

$$=tan^{-1}(infty)-tan^{-1}(1)+tan^{-1}(infty)-tan^{-1}(0)=$$

$$=pi-frac{pi}{4}=frac{3pi}{4}.$$

share|cite|improve this answer

answered Jan 31 at 2:21

DXTDXT

5,8892732

$endgroup$

$$sum^{infty}_{n=1}bigg(tan^{-1}(n+1)-tan^{-1}(n)bigg)$$
$$+sum^{infty}_{n=1}bigg(tan^{-1}(n)-tan^{-1}(n-1)bigg)$$

Convert into Telescopic Series

$$=tan^{-1}(infty)-tan^{-1}(1)+tan^{-1}(infty)-tan^{-1}(0)=$$

$$=pi-frac{pi}{4}=frac{3pi}{4}.$$

share|cite|improve this answer

answered Jan 31 at 2:21

DXTDXT

5,8892732

share|cite|improve this answer

share|cite|improve this answer

answered Jan 31 at 2:21

DXTDXT

5,8892732

answered Jan 31 at 2:21

DXTDXT

5,8892732

answered Jan 31 at 2:21

DXTDXT

5,8892732

5,8892732

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.

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

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3094394%2fevaluate-sum-n-1-infty-arctan-frac2n2%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

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

Name

Email

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

Name

Email

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

Name

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Name

Name

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

Email

Required, but never shown

This page is only for reference, If you need detailed information, please check here