Limit of the finite series $\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$ The problem is to find the limit of:
$$\ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}$$
A the series is finite, it looks as if it would be required to find the sum of the series - however, I have to find the limit. It resembles for me:
$$\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n^2-kn }{(k+n)^3-2kn^2-2k^2n}=\\
\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)^3-2kn(n+k)}=\\
\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ (k+n)^2+n(n-k) }{(k+n)((k+n)^2-2kn)}$$
but I don't know what to do next and how to solve it. I would appreciate your help.
Edit: Does it has something in common with Riemann sum?

Attempts
Using the integral test, we can show the  convergence since
$$I=\int  \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3}\,dk=\int \left(\frac{n}{k^2+n^2}+\frac{1}{k+n}\right)\,dk=$$
$$I=\log (k+n)-\tan ^{-1}\left(\frac{n}{k}\right)$$
Integrating between $k=1$ and $k=\lfloor n + \sqrt{n}\rfloor$ and simplifying, we have
$$\tan ^{-1}\left(\frac{n \left(\left\lfloor n+\sqrt{n}\right\rfloor
   -1\right)}{\left\lfloor n+\sqrt{n}\right\rfloor +n^2}\right)+\log \left(\frac{\left\lfloor n+\sqrt{n}\right\rfloor +n}{n+1}\right)$$ which is asymptotic to $\frac \pi 4+\log(2)\approx 1.47855$.
Integrating between $k=0$ and $k=\lfloor n + \sqrt{n}\rfloor$ and simplifying, we have
$$\log \left(\frac{\left\lfloor n+\sqrt{n}\right\rfloor +n}{n}\right)-\tan
   ^{-1}\left(\frac{n}{\left\lfloor n+\sqrt{n}\right\rfloor }\right)+\frac{\pi }{2}$$ which shows the same asymptotic value.
On the other hand, we can write the summation as
$$-\frac{1}{2} i H_{\left\lfloor n+\sqrt{n}\right\rfloor -i n}+\frac{1}{2} i H_{i
   n+\left\lfloor n+\sqrt{n}\right\rfloor }+H_{n+\left\lfloor n+\sqrt{n}\right\rfloor
   }-H_n-\frac{1}{2 n}+\frac{1}{2} \pi  \coth (\pi  n)$$ and using the asymptotics of harmonic numbers, the same result.
