$\lim_{n \to \infty} {\frac{1}{n^2}\sum_{k=0}^{n}{\frac{1}{\ln{(1 + \frac{(n+k)\sqrt{n^2+k^2}}{n^3})}}}}$

Question

How can I solve

$$\lim_{n \to \infty} {\frac{1}{n^2}\sum_{k=0}^{n}{\frac{1}{\ln{(1 + \frac{(n+k)\sqrt{n^2+k^2}}{n^3})}}}}$$

It looks like a Riemann limit to me, but I'm not able to get it to a final form.

Answer 1

$$\lim_{n \to \infty} {\frac{1}{n^2}\sum_{k=0}^{n}{\frac{1}{\log{(1 + \frac{(n+k)\sqrt{n^2+k^2}}{n^3})}}}} = \lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^n\frac1{\log(1+\frac1n(1+\frac kn)\sqrt{1+\frac{k^2}{n^2}})}.$$

Now, remembering that $x \geq \log(1+x)$ for every $x>0$ and that for every $\varepsilon>0$ there is a $\delta>0$ such that for every $x \in (0,\delta)$ we have $\log(1+x) \geq (1-\varepsilon)x$.

The first inequality implies $$\lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^n\frac1{\log(1+\frac1n(1+\frac kn)\sqrt{1+\frac{k^2}{n^2}})} \geq \lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^n\frac1{\frac1n(1+\frac kn)\sqrt{1+\frac{k^2}{n^2}}} =$$ $$= \lim_{n\to\infty}\frac1{n}\sum_{k=0}^n\frac1{(1+\frac kn)\sqrt{1+\frac{k^2}{n^2}}} = \int_0^1\frac1{(1+x)\sqrt{1+x^2}}dx.$$

On the other hand, since (for a fixed $\varepsilon$) for $n$ big enough we have that for every $k$ it holds that $\frac1n(1+\frac kn)\sqrt{1+\frac{k^2}{n^2}} \leq \frac{2\sqrt2}n \leq \delta$, $$\lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^n\frac1{\log(1+\frac1n(1+\frac kn)\sqrt{1+\frac{k^2}{n^2}})} \leq \frac1{1-\varepsilon}\int_0^1\frac1{(1+x)\sqrt{1+x^2}}dx.$$

Since $\varepsilon$ is arbitrary, we get $$\lim_{n\to\infty}\frac1{n^2}\sum_{k=0}^n\frac1{\log(1+\frac1n(1+\frac kn)\sqrt{1+\frac{k^2}{n^2}})} =\int_0^1\frac1{(1+x)\sqrt{1+x^2}}dx.$$