Ideas Needed !

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 :

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

#Calculus #Summation #Ideas #Roopesh #Outoftags

Easy Math Editor

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Comments

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

Raghav Vaidyanathan - 6 years, 3 months ago

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

A Former Brilliant Member - 6 years, 3 months ago

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

A Former Brilliant Member - 6 years, 3 months ago

@Akshay Bodhare So you are back ?

A Former Brilliant Member - 6 years, 3 months ago

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,

$S \approx 0.7345725812$

Akshay Bodhare - 6 years, 2 months ago

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

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

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – Could you please give me some hint(not solution) for this? I want to solve it!But if you think that I am violating any guidelines of Brilliant , you can delete this comment.

Harsh Shrivastava - 6 years, 2 months ago

@Harsh Shrivastava – I'm sure you must have heard about Cauchy Schwartz inequality , haven't you ?

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

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – Hmmmm... I will try using it,

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

Harsh Shrivastava - 6 years, 2 months ago

@Harsh Shrivastava – No problem :)

A Former Brilliant Member - 6 years, 2 months ago

@A Former Brilliant Member – However,even the upper bound I gave is not tight enough,do you have any ideas for better approximation?

Akshay Bodhare - 6 years, 2 months ago

@Akshay Bodhare – I think that I'll have to search on the I'net for it then , since my knowledge on it is limited .

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

A Former Brilliant Member - 6 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$