Answer to Question #308445 in Real Analysis for Pankaj

Question

Evaluate:

Lim↓ n→∞ [ n/1+n² + n/4+ n² +n/9+ n² +...... +n/2n²]

Answer 1

ANSWER : $\frac{\pi}{4}$

EXPLANATION

We transform the sum $\frac{n}{1+n^2}+\frac{n}{4+n^2}+\frac{n}{9+n^2}+ \cdots+\frac{n}{2n^2}$ as follows :$\frac{1 }{n}\left ( \frac{1}{\left (\frac{1}{n} \right ) ^{2} +1 }+\frac{1}{\left (\frac{2}{n} \right ) ^{2} +1 }+\cdots+\frac{1}{\left (\frac{n}{n} \right ) ^{2} +1 } \right )$ = $\sum _{k=1}^{n}\frac{1}{n}\cdot \frac{1}{\left (\frac{k}{n} \right ) ^{2} +1 }.$ The resulting sum is the Riemann integral sum for the function $f(x)=\frac{1}{1+x^2}$ on the segment [0,1] divided into $n$ equal parts by points $x_{k}=\frac{k}{n}\left ( k=0,\cdots\,n \right )$ . Since $x_k-x_{k-1}=\frac{1}{n}$ for $k=1,\cdots,\,n$ then $\sum _{k=1}^{n}\frac{1}{n}\cdot \frac{1}{\left (\frac{k}{n} \right ) ^{2} +1 } =\sum _{k=1}^{n}\ \left ( x_k-x_{k-1} \right ) \cdot \ f(x_k)$ .

The function $f$ is integrable, because $f$ is continuous. Therefore $\lim _{n\rightarrow\infty}\sum _{k=1}^{n}\frac{1}{n}\cdot \frac{1}{\left (\frac{k}{n} \right ) ^{2} +1 } =\lim_{n\rightarrow\infty}\sum _{k=1}^{n}\ \left ( x_k-x_{k-1} \right ) \cdot \ f(x_k)=\int_{0}^{1}f(x)dx$ .

$\int_ {0}^{1}\frac{1}{1+x ^{2}}dx=\left [ \arctan 1-\arctan 0 \right ] =\frac{ \pi }{4}$ . Hence $\lim _{n\rightarrow\infty}\sum _{k=1}^{n}\frac{1}{n}\cdot \frac{1}{\left (\frac{k}{n} \right ) ^{2} +1 } = \frac{\pi}{4}$ .