Question

$I{t}_{n\to \infty }\{\frac{1}{1+n^2}+\frac{2}{2+n^2}+\frac{3}{3+n^2}+{\cdots }_{\cdots \cdots }+\frac{n}{n+n^2}\}$

• A. 1
• B. $\frac{1}{2}$
• C. $\infty$

Answer 1

The correct option is B $\frac{1}{2}$

Let $f\left(x\right)=\stackrel{lt}{n\to \infty }\frac{1}{1+n^2}+\frac{2}{1+n^2}+\cdots \cdots \cdots +\frac{n}{1+n^2}$
$=\stackrel{lt}{n\to \infty }\frac{n(n+1)}{2(1+n^2)}=\frac{1}{2}$
Let $g\left(x\right)=\stackrel{lt}{n\to \infty }\frac{1}{n+n^2}+\frac{2}{n+n^2}+\cdots \cdots \cdots +\frac{n}{n+n^2}$
$=\stackrel{lt}{n\to \infty }\frac{n(n+1)}{2(1+n^2)}=\frac{1}{2}$
We know that $f\left(x\right)\ge Q\ge g\left(x\right)$ where Q = $\stackrel{lt}{n\to \infty }\frac{1}{n+n^2}+\frac{2}{n+n^2}\cdots \cdots +\frac{n}{n+n^2}$
$\therefore Q=\frac{1}{2}$