Is there a better way to find the following sum ? I mean is there a better approximation possible ?

\[\sum_{n=1}^{\infty} \sqrt {1+n^4} -n^2\]

A possible way could be :

\[ S = \displaystyle \sum_{n=1}^{\infty} \sqrt{1 + n^{4}} - n^{2} \\ S = \displaystyle \sum_{n=1}^{\infty} \frac { \left ( \sqrt{1 + n^{4}} - n^{2} \right ) \left ( \sqrt{1 + n^{4}} + n^{2} \right ) }{ \sqrt{1 + n^{4}} + n^{2} } \\= \displaystyle \sum_{n=1}^{\infty} \frac { 1 } { \sqrt{1 + n^{4}} + n^{2} } \\\\ \text{Now we can make an approximation that } \\\\S < \displaystyle \sum_{n=1}^{\infty} \frac { 1 } { \sqrt{ n^{4}} + n^{2} } \\ S < \displaystyle \sum_{n=1}^{\infty} \frac { 1 } { 2 n^{2} } \\ S < \displaystyle {\frac { \pi^{2}}{12} } \]

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## Comments

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TopNewest@Akshay Bodhare So you are back ?

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Sorry for replying after a long time,

\(\frac{\pi^2}{12} \approx 0.8224\)

This bound is not tight enough,I suggest that you should use Cauchy-Schwarz inequality in last step to get the better approximation,

\(S\approx\dfrac{\pi \coth {\frac{\pi}{\sqrt{1+\sqrt{2}}}}-\sqrt{1+\sqrt{2}}}{\sqrt{2(1+\sqrt{2})}}\approx 0.7736211057\)

Btw,the sum coverges slowly

The approximate value of the sum according to CAS is,

\(S \approx 0.7345725812\)

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Nice , I wonder why it didn't strike me ! +1

Also I'll try to think up a question based on it .

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I'll discuss with some of my other friends and I'll definitely get back to you :)

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this? I want to solve it!But if you think that I am violating any guidelines of Brilliant , you can delete this comment.

Could you please give me some hint(not solution) forLog in to reply

Also I can't delete any comment , I'm not a moderator you see !

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I thought you were a moderator, Sorry to bother you.

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@Brian Charlesworth sir,@Kartik Sharma ,@Pratik Shastri ,@Ronak Agarwal ,@Vraj Mehta

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If you just want a better approximation, just calculate the sum of first few terms by hand, and use approximation for the rest.

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I guess that'll do it but ok I'll try

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