<s>[INST] 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?
 [/INST] 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.
</s>