Question

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

Answer 1

This problem uses sandwich theorem.

The given series is ${lim}_{n\to \infty }{\sum }_{r=1}^n\frac{r}{n^2+n+r}$ which can be expressed as ${lim}_{n\to \infty }{S}_n$ where $S_n={\sum }_{r=1}^n\frac{r}{n^2+n+r}$.

Now, we can see that

$\frac{r}{n^2+n}\le \frac{r}{n^2+n+r}\le \frac{r}{n^2+2n}$

Applying ${\sum }_{r=1}^n$ on the inequality, we get

$\Rightarrow {\sum }_{r=1}^n\frac{r}{n^2+n}\le {\sum }_{r=1}^n\frac{r}{n^2+n+r}\le {\sum }_{r=1}^n\frac{r}{n^2+2n}\Rightarrow \frac{1}{n^2+n}{\sum }_{r=1}^{n}r\le S_n\le \frac{1}{n^2+2n}{\sum }_{r=1}^{n}r\Rightarrow \frac{1}{n(n+1)}\frac{n(n+1)}{2}\le S_n\le \frac{1}{n(n+2)}\frac{n(n+1)}{2}\Rightarrow \frac{1}{2}\le S_n\le \frac{1}{2}\frac{(n+1)}{(n+2)}$

Taking ${lim}_{n\to \infty }$ on the inequality, we get

$\Rightarrow {lim}_{n\to \infty }\frac{1}{2}\le {lim}_{n\to \infty }{S}_n\le {lim}_{n\to \infty }\frac{1}{2}\frac{(n+1)}{(n+2)}\Rightarrow \frac{1}{2}\le {lim}_{n\to \infty }{S}_n\le \frac{1}{2}.{lim}_{n\to \infty }\frac{(1+\frac{1}{n})}{(1+\frac{2}{n})}\Rightarrow \frac{1}{2}\le {lim}_{n\to \infty }{S}_n\le \frac{1}{2}.\frac{1+0}{1+0}\Rightarrow \frac{1}{2}\le {lim}_{n\to \infty }{S}_n\le \frac{1}{2}\Rightarrow {lim}_{n\to \infty }{S}_n=\frac{1}{2}.$

Hence the required limit is $\frac{1}{2}$