<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] 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:
$$ \lim_{n\to\infty}\sum_{k=1}^{\lfloor n+\sqrt{n}\, \rfloor} (\frac{n}{k^2+n^2}+\frac{1}{k+n}) = \frac{\pi}{4}+\ln(2)$$
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