Question

The value of ${lim}_{n\to \infty }{\sum }_{r=1}^n\frac{r^2}{n^3+n^2+r}$ equals

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

Answer 1

Answer: (A)

$\frac{1}{3}$

This problem uses sandwich theorem.

The given series is ${lim}_{n\to \infty }{\sum }_{r=1}^n\frac{r^2}{n^3+n^2+r}$ which can be expressed as ${lim}_{n\to \infty }{S}_n$ where $S_n={\sum }_{r=1}^n\frac{r^2}{n^3+n^2+r}$.

Now, we can see that

$\frac{r^2}{n^3+n^2+n}\le \frac{r^2}{n^3+n^2+r}\le \frac{r^2}{n^3}$

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

$\Rightarrow {\sum }_{r=1}^n\frac{r^2}{n^3+n^2+n}\le {\sum }_{r=1}^n\frac{r^2}{n^3+n^2+r}\le {\sum }_{r=1}^n\frac{r^2}{n^3}\Rightarrow \frac{1}{n^3+n^2+n}{\sum }_{r=1}^n{r}^2\le S_n\le \frac{1}{n^3}{\sum }_{r=1}^n{r}^2\Rightarrow \frac{1}{n^3+n^2+n}\frac{n(n+1)(2n+1)}{6}\le S_n\le \frac{1}{n^3}\frac{n(n+1)(2n+1)}{6}\Rightarrow \frac{1}{6}\frac{n(n+1)(2n+1)}{n^3+n^2+n}\le S_n\le \frac{1}{6}\frac{n(n+1)(2n+1)}{n^3}$

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

$\Rightarrow {lim}_{n\to \infty }\frac{1}{6}\frac{n(n+1)(2n+1)}{n^3+n^2+n}\le {lim}_{n\to \infty }{S}_n\le {lim}_{n\to \infty }\frac{1}{6}\frac{n(n+1)(2n+1)}{n^3}\Rightarrow {lim}_{n\to \infty }\frac{1}{6}\frac{{2n}^3+3n^2+n}{n^3+n^2+n}\le {lim}_{n\to \infty }{S}_n\le {lim}_{n\to \infty }\frac{1}{6}\frac{{2n}^3+3n^2+n}{n^3}\Rightarrow \frac{1}{6}.{lim}_{n\to \infty }\frac{(2+\frac{3}{n}+\frac{1}{n^2})}{(1+\frac{1}{n}+\frac{1}{n^2})}\le {lim}_{n\to \infty }{S}_n\le \frac{1}{6}.{lim}_{n\to \infty }\frac{(2+\frac{3}{n}+\frac{1}{n^2})}{1}\Rightarrow \frac{1}{6}\frac{(2+0+0)}{(1+0+0)}\le {lim}_{n\to \infty }{S}_n\le \frac{1}{6}\frac{(2+0+0)}{1}\Rightarrow \frac{1}{3}\le {lim}_{n\to \infty }{S}_n\le \frac{1}{3}\Rightarrow {lim}_{n\to \infty }{S}_n=\frac{1}{3}.$

Hence the required limit is $\frac{1}{3}.$

Therefore, Option $A$ is correct.