1 votes 1 votes 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 $\frac{\pi}{4}$ $\frac{\pi}{8}$ $\frac{\pi}{6}$ $2 \pi$ Quantitative Aptitude isi2014-dcg quantitative-aptitude summation non-gate + – Arjun asked Sep 23, 2019 • recategorized Nov 12, 2019 by Lakshman Bhaiya Arjun 489 views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments techbd123 commented Nov 3, 2019 reply Follow Share 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. 0 votes 0 votes Lakshman Bhaiya commented Nov 3, 2019 reply Follow Share $\displaystyle \sum_{k=0}^{\infty}\frac{n}{n^2+k^2}$ ? $$\mathrm{Or}$$ $\displaystyle \frac{n}{n^2}+\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\cdots+\frac{n}{n^2+(n-1)^2}+\frac{n}{n^2+n^2}+\frac{n}{n^2+(n+1)^2}+\frac{n}{n^2+(n+2)^2}+\cdots$ These are same. 1 votes 1 votes techbd123 commented Nov 3, 2019 reply Follow Share 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 $\displaystyle \lim_{n \to \infty} \left( \dfrac{n}{n^2}+\dfrac{n}{n^2+1^2}+\dfrac{n}{n^2+2^2}+ \cdots + \dfrac{n}{n^2+(n-1)^2} +\dfrac{n}{n^2+n^2} \right)$ 0 votes 0 votes Please log in or register to add a comment.
