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

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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 \(n\).

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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

Shivang Jindal Ronak Agarwal Pratik Shastri Sudeep Salgia @Rube

What is p?

sorry its \(\pi\)

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`\sin \theta`

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## Comments

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TopNewestCan 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 \(n\).

Log in to reply

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

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Shivang Jindal Ronak Agarwal Pratik Shastri Sudeep Salgia @Rube

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What is p?

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sorry its \(\pi\)

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