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$\displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^2} \cos\Big(\dfrac{9}{n\pi + \sqrt{(n\pi)^2 - 9}}\Big) = - \dfrac{\pi^2}{12e^3}$

Easy Math Editor

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Comments

Can you approximate $n \pi - \sqrt{ ( n \pi ) ^2 - 9 }$ as $n$ gets large? Use Big-O notation if you are familiar with that.

Are you sure that there is an $e^3$ in there?

Also, please use distinct values for your indices. You cannot take the sum of the variable $n$ , as it goes from 1 to $n$ .

Calvin Lin Staff - 6 years, 5 months ago

I am not familiar with that , I am sure that this is the question . Unable to solve that's why shared it and mentioned some brilliant members.

In a magazine - Mathematics today I saw it , Chapter - definite integration , section - Sandwich theorem

U Z - 6 years, 5 months ago

Shivang Jindal Ronak Agarwal Pratik Shastri Sudeep Salgia @Rube

U Z - 6 years, 5 months ago

What is p?

Abhay Agnihotri - 6 years, 5 months ago

sorry its $\pi$

U Z - 6 years, 5 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}`	$\boxed{123}$