$\lim _{n \rightarrow \infty}\left(\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\frac{n}{n^{2}+3^{2}}

Question

$\lim _{n ightarrow \infty}\left(\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\frac{n}{n^{2}+3^{2}}+\ldots .+\frac{1}{5 n}ight)$ is equal to\

Professor PC Thomas Classes · Answer

(a) : Given,\ $\lim _{n ightarrow \infty}\left(\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\ldots+\frac{1}{5 n}ight)$\ $=\lim _{n ightarrow \infty} \sum_{r=1}^{2 n} \frac{n}{n^{2}+r^{2}}=\lim _{n ightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n\left(1+\frac{r^{2}}{n^{2}}ight)}$\ $=\int_{0}^{2}\left(\frac{1}{1+x^{2}}ight) d x$\ [Putting $\frac{r}{n}=x$, we get $\frac{1}{n} d r=d x$ and\ $\qquad r=1 \Rightarrow x=0, r=2 n \Rightarrow x=2]$\ $=\left[	an ^{-1} xight]_{0}^{2}=	an ^{-1} 2$\