A problem from Makarov Selected Problems in Real Analysis

Question

This is a problem from Makarov’s Selected Problems in Real Analysis. Put $S(n) = \sum_{k=0}^{2n}{\frac{k}{k+n^2}}$ Find the limit as n tends to infinity. The answer is $\frac {1}{2} \ln 5$. My solution goes as follows. Since $y=\frac{x}{x+n^2}$ is increasing on $[0, \infty)$ , we have $$\int_{0}^{2n}\frac{x dx}{x+n^2}\leq S(n) \leq \int_{0}^{2n+1}\frac{x dx}{x+n^2}$$ Integrating both sides and letting n tend to infinity we get the limit 2. (LHS becomes $2n+n^2\log \frac{n}{n+2}$ and RHS $2n+1+2n^2\log{\frac{n}{2n+1}}$) Could you please point out what is wrong with my answer?

Answer 1

It is true that

$\sum\limits_{k=0}^{2n}{\frac{k}{k^2+n^2}}\lt S(n)\lt \sum\limits_{k=0}^{2n}{\frac{k}{n^2}}$

Both sides can be formed to Riemann integrals:

$2\frac{1}{2n}\sum\limits_{k=0}^{2n}{\frac{2\frac{k}{2n}}{4\big(\frac{k}{2n}\big)^2+1}}\lt S(n)\lt 4\frac{1}{2n} \sum\limits_{k=0}^{2n}{\frac{k}{2n}}$

$\frac{1}{2}\int\limits_0^1 \frac{8x}{4x^2+1}dx\lt S(n)\lt 4 \int\limits_0^1xdx$

$\frac{1}{2}ln({4x^2+1})\lt S(n)\lt 4 \frac{x^2}{2}$ between $0$ and $1$.

So I got the following bounds: $\frac{1}{2}ln{5}\lt S(n)\lt 2$