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# How to start?

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

Note by Megh Choksi
2 years, 4 months ago

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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 $$n$$. Staff · 2 years, 4 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 · 2 years, 4 months ago

Shivang Jindal Ronak Agarwal Pratik Shastri Sudeep Salgia @Rube · 2 years, 4 months ago

What is p? · 2 years, 4 months ago

sorry its $$\pi$$ · 2 years, 4 months ago