Evaluating the limit $\lim_{n\to\infty} \left[ \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$ Evaluating the limit $\displaystyle \lim_{n\to\infty} \left[  \frac{n}{n^2+1}+ \frac{n}{n^2+2^2} + \ldots + \frac{n}{n^2+n^2} \right]$
I have a question about the following solution:
We may write it in the form:
$$ \frac{1}{n} \left[ \frac{1}{1+(\frac{1}{n})^2} + \ldots + \frac{1}{1+(\frac{n}{n})^2} \right] $$
Somehow I need to figure out that the limit is actually the Riemann sum of $\frac{1}{1 + x^2}$ on $[0,1]$ for $\pi = 0 < \frac{1}{n} < \ldots < \frac{n}{n}$.
Can you explain to me to reach this conclusion? 

Hints : First look at the coefficient $\frac{1}{n}$ in front of the expression inside brackets. The $1$ at the top of the fraction is $1-0$, where $1$ and $0$ are the end points of the interval $[0,1]$. The $n$ at the bottom would be the length of a subinterval of $[0,1]$ when you partition $[0,1]$ into $n$ equal parts.
Then look at each individual term inside brackets. For example,
$$
\frac{1}{1+(\frac{3}{n})^2}
$$
would be the value of the function $\frac{1}{1+x^2}$ when $x$ is $3$ times the length of a subinterval of length $\frac{1}{n}$ added from the starting point $0$.
Can you now figure out that this is the Riemann sum you had to recognize ? What the expression really represents is a sum of areas of rectangles of base $\frac{1}{n}$ and height $\frac{1}{1+(\frac{i}{n})^2}$. Here you approach the area under the curve $\frac{1}{1+x^2}$ over $[0,1]$ by taking the right end points of the subintervals of an equally distributed partition.
