ISI2014-DCG-65

Question

The sum $\dfrac{n}{n^2}+\dfrac{n}{n^2+1^2}+\dfrac{n}{n^2+2^2}+ \cdots + \dfrac{n}{n^2+(n-1)^2} + \cdots \cdots$ is

• A. $\frac{\pi}{4}$
• B. $\frac{\pi}{8}$
• C. $\frac{\pi}{6}$
• D. $2 \pi$

Comments

Nope. Both aren't the same.

$\displaystyle \lim_{n\to \infty} \sum_{k=0}^{n}\frac{n}{n^2+k^2}$ has the last term as $\frac{n}{n^2+n^2}$ where $n\to \infty$.

 

But $\frac{n}{n^2}+\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\cdots+\frac{n}{n^2+(n-1)^2}+\frac{n}{n^2+n^2}+\frac{n}{n^2+(n+1)^2}+\frac{n}{n^2+(n+2)^2}+\cdots$ can have term $\frac{n}{n^2+(n+n)^2}=\frac{n}{n^2+4n^2}$ and so on.

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$\displaystyle \sum_{k=0}^{\infty}\frac{n}{n^2+k^2}$ ?
$$\mathrm{Or}$$
$\displaystyle \frac{n}{n^2}+\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\cdots+\frac{n}{n^2+(n-1)^2}+\frac{n}{n^2+n^2}+\frac{n}{n^2+(n+1)^2}+\frac{n}{n^2+(n+2)^2}+\cdots$

These are same.

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If so, the answer would be $\frac{\pi}{2}$ which is NOT there in the stated options.

BTW $\displaystyle \lim_{n\to \infty} \sum_{k=0}^{n}\frac{n}{n^2+k^2}=\frac{\pi}{4}$

but $\displaystyle  \sum_{k=0}^{\infty}\frac{n}{n^2+k^2}=\frac{\pi}{2}$

 

So definitely the question has to be
$\displaystyle \lim_{n \to \infty} \left( \dfrac{n}{n^2}+\dfrac{n}{n^2+1^2}+\dfrac{n}{n^2+2^2}+ \cdots + \dfrac{n}{n^2+(n-1)^2} +\dfrac{n}{n^2+n^2} \right)$