# Evaluate the following limit: $\lim_{n\to\infty}\left(\sum_{r=1}^{n}{\frac{r}{n^{2}+n+r}}\right)$

Evaluate the following limit:

$$\lim_{n\to\infty}\left(\sum_{r=1}^{n}{\frac{r}{n^{2}+n+r}}\right)$$

The answer given is $\frac{1}{2}$.

Squeeze it: $$\frac{r}{n^2+n+n}\leq\frac{r}{n^2+n+r}\leq\frac{r}{n^2+n}$$
• @alex.jordan The hints says, for all $n$, that $$\sum_{r=1}^n \frac{r}{n^2+n+n} <leq \sum_{r=1}^n \frac{r}{n^2+n+r} <leq \sum_{r=1}^n \frac{r}{n^2+n}$$ Now consider what happens when $n$ grows large. What happens to the difference between the upper limit and lower limit? – Sasha Sep 8 '12 at 16:28
Yes I have got it thank you. First put the expression as $\frac{r}{n^2+n+n}$ and then evaluate the limit. It will become $\frac{1}{2}$. Now put the given expression as $\frac{r}{n^2+n}$. Again the value will come out to be $\frac{1}{2}$. So from Sandwich Theorem we will have the limit as $\frac{1}{2}$. Thanks again