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







share|cite|improve this question











$endgroup$








  • 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
















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







share|cite|improve this question











$endgroup$








  • 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














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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 13 '17 at 7:14









Sangchul Lee

93.6k12167271




93.6k12167271










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














  • 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










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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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













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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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


















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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    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
















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.






share|cite|improve this answer









$endgroup$



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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Sep 13 '17 at 7:21









Lord Shark the UnknownLord Shark the Unknown

104k1160132




104k1160132












  • $begingroup$
    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










  • $begingroup$
    @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




















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