Find the limit $\lim_{n\to \infty}\sum_{i=1}^{n}\frac{i}{n^2+i^2}$ by expressing it as a definite integral of an appropiate function via Riemann Sums.
Observation: $n$ must refer to the number of slices, and $i$ must refer to $i$th slice.
My attempt.
First I revisited Riemann Sums. Assume what I am trying to find have the form $\int_{a}^{b}f(x).$ Cutting up the bound $(a,b)$ into $n$ slices, the length of each piece with respect to $x$ is $\frac{b-a}{n}. $
Next, looking at each slice, I decide to take the right hand value for convenience, that is $a+i\frac{b-a}{n}. $ Now, I clearly have the area of each slice, that is, $\frac{b-a}{n}*f(a+i\frac{b-a}{n} )$
And when I sum up all of the slices, I have $\sum_{i=1}^{n}\frac{b-a}{n}*f(a+i\frac{b-a}{n} )$
And finally, increasing the number of cuts to make the area as accurate as possible, we have $\lim_{n\to \infty}\sum_{i=1}^{n}\frac{b-a}{n}*f(a+i\frac{b-a}{n} )$
Therefore I conclude that $\frac{i}{n^2+i^2}=\frac{b-a}{n}*f(a+i\frac{b-a}{n} )$
Whats left now is to find b,a and f(x). After all the work, I feel closer to my answer, yet so far away from it.
Any hints? Thanks in advance! List them as solutions. I am looking for hints only.