Evaluate the following limit:
$$\lim_{n\to\infty}\left(\sum_{r=1}^{n}{\frac{r}{n^{2}+n+r}}\right)$$
The answer given is $\frac{1}{2}$.
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: $$ \frac{r}{n^2+n+n}\leq\frac{r}{n^2+n+r}\leq\frac{r}{n^2+n} $$
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