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?
 A: First, we can omit the $\sqrt{n}$ part and only sum up to $n$. To see this,
$$
\sum_{k=n}^{n+\sqrt{n}} x_n \leq 
\sum_{k=n}^{n+\sqrt{n}} \frac{4(n+\sqrt{n})^2}{4n^3} \leq \sum_{k=n}^{n+\sqrt{n}} \frac{(n+n)^2}{n^3} \leq 
\sum_{k=n}^{n+\sqrt{n}} \frac{4}{n} \leq 4\frac{\sqrt{n}}{n} \mapsto 0
$$
The rest is the Riemannian sum of the integral
$$
\int_0^1 \frac{x^2 + x + 2}{x^3 + x^2 + x + 1} \mathrm{d}x = \int_0^1 \frac{1}{1+x} \mathrm{d}x + \int_0^1 \frac{1}{1+x^2}\mathrm{d}x.
$$
This can be easily computed using calculus.
A: 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.
A: Thanks to Peter Franek I would try to calculate the limit.
The sum can be rewritten as:
$$\ \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} \frac{ k^2+kn+2n^2 }{k^3+k^2n+kn^2+n^3} = \sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} (\frac{n}{k^2+n^2}+\frac{1}{k+n}) $$
$$\ a_n=\frac{n}{k^2+n^2} = \frac{n}{n^2}\cdot\frac{1}{\frac{k^2}{n^2}+1}$$
$$\ b_n=\frac{1}{k+n} = \frac{1}{n}\cdot\frac{1}{\frac{k}{n}+1} $$
Let us consider $f(x) = \frac{1}{x^2+1} $ and $g(x)=\frac{1}{x+1}$
$$\lim_{n\to\infty}a_n=\int_0^1f(x)dx=\int_0^1\frac{1}{x^2+1}dx=\arctan(x)|_0^1=\frac{\pi}{4} \\\lim_{n\to\infty}b_n=\int_0^1g(x)dx=\int_0^1\frac{1}{x+1}dx = \int_0^1\frac{c'(x)}{c(x)}=\ln(x+1)|_0^1=\ln(2)$$
As the element $\sqrt(n)$ becomes unsignificant for $n\to\infty$ as $\frac{\sqrt(n)}{n}\to0$, the limit is:
$$ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} (\frac{n}{k^2+n^2}+\frac{1}{k+n}) = \frac{\pi}{4}+\ln(2)$$
