Question

$\underset{n\to \infty }{lim}\sum _{r=1}^n\frac{r}{n^2+n+r}$ is equal to

• A. $0$
• B. $\frac{1}{3}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

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

Let $f\left(n\right)=\frac{1}{n^2+n+1}+\frac{2}{n^2+n+2}+\cdots +\frac{n}{n^2+n+n}$
Consider $g\left(n\right)=\frac{1}{n^2+n+n}+\frac{2}{n^2+n+n}+\cdots +\frac{n}{n^2+n+n}$
$=\frac{1+2+3+\cdots +n}{n^2+2n}=\frac{n(n+1)}{2(n^2+2n)}$
where $g\left(n\right)<f\left(n\right)\dots \left(1\right)$

Similarly, $h\left(n\right)=\frac{1}{n^2+n+1}+\frac{2}{n^2+n+1}+\cdots +\frac{n}{n^2+n+1}$
$=\frac{n(n+1)}{2(n^2+n+1)}$
where $f\left(n\right)<h\left(n\right)\dots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$
$g\left(n\right)<f\left(n\right)<h\left(n\right)$

But $\underset{n\to \infty }{lim}g\left(n\right)=\underset{n\to \infty }{lim}h\left(n\right)=\frac{1}{2}(\because \text{We know, as}n\to \infty ,\frac{1}{n}\to 0)$
Hence, using Sandwich theorem, $\underset{n\to \infty }{lim}f\left(n\right)=\frac{1}{2}$