Find $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$ 
Find $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$

Since $\frac{k}{n^2+k^2}\leq \frac{k}{k^2+k^2}=\frac{1}{2k}$, then $\sum_{k=1}^n \frac{k}{n^2+k^2}\leq \sum_{k=1}^n \frac{1}{2k}=\frac{1}{2} \sum_{k=1}^n\frac{1}{k}$.
Now we send $n$ to infinity, then since $\sum_{k=1}^\infty \frac{1}{k}$ is harmonic, $\lim_{n\to \infty} \sum_{k=1}^n \frac{k}{n^2+k^2}$ doesn't exist.
I wonder if my thinking is right.

Just for the fun.
Consider $$S_n= \sum_{k=1}^n \frac{k}{n^2+k^2}=\sum_{k=1}^n \frac{k}{(n^2+ik)(n-ik)}=\frac i2  \sum_{k=1}^n\left(\frac{1}{n+i k}-\frac{1}{n-i k} \right)$$ Using generalized harmonic numbers $$S_n=\frac{1}{2} \left(-H_{-i n}-H_{i n}+H_{(1-i) n}+H_{(1+i) n}\right)$$ Using asymptotics $$H_p=\gamma +\log \left({p}\right)+\frac{1}{2 p}-\frac{1}{12
   p^2}+O\left(\frac{1}{p^4}\right)$$ and applying to each term, after simplifications, you should end with $$S_n=\frac{\log (2)}{2}+\frac{1}{4 n}-\frac{1}{12
   n^2}+O\left(\frac{1}{n^4}\right)$$ which shows the limit and how it is approached.
Use it with $n=10$ which gives $$S_{10}=\frac{2892380100711541}{7801656832544900}\approx 0.370739$$ while the expansion gives $$S_{10}\approx\frac{\log (2)}{2}+\frac{193}{8400}\approx 0.369550$$
