Evaluate $\lim_{n\to {\infty}} \sum_{j=1}^n \frac{j}{n^2 +j^2}$

Question

The limit in question is $$\lim_{n\to {\infty}} \sum_{j=1}^n \frac{j}{n^2 +j^2}$$

and I'm trying to approach it via Riemann sums. I think a partition can be chosen ( although I'm not entirely sure why ), such that the length of each sub-interval $\Delta x_i=\frac{j-(j-1)}{n}=\frac{1}{n}$. Then choose point $c_i=\frac{j}{n}\in \Delta x_i$. Now $j=c_i \cdot n$. And the sum can be rewritten as $$\sum_{j=1}^n \frac{c_in}{n^2(1+c_i^2)}=\sum_{j=1}^n \frac{c_i}{n(1+c_i^2)}$$

Which is obviously wrong since there's a $n$ in the denominator. Anyway, I think the point here is to apply $$\lim_{n\to {\infty}} (\sum_{j=1}^n f(c_i)\Delta x_i) = \int_a^b f(x) dx$$

And then $b=1$, $a=0$. Because of $$0\leq \frac{j}{n^2 +j^2} \leq 1$$ However, I think I have not understood the concept of Riemann sums very well and would like some help.

Answer 1

We have $$ \sum\limits_{j = 1}^n {\frac{j}{{n^2 + j^2 }}} = \frac{1}{n}\sum\limits_{j = 1}^n {\frac{{j/n}}{{1 + (j/n)^2 }}} \to \int_0^1 {\frac{x}{{1 + x^2 }}dx}=\frac{1}{2}\log 2 . $$

Answer 2

$$\sum_{k=1}^n\frac k{n^2+k^2}=\frac1n\sum_{k=1}^n\frac{\frac kn}{1+\left(\frac kn\right)^2}\xrightarrow[n\to\infty]{}\int_0^1\frac x{1+x^2}dx=\ldots$$