Compute the following limit, possibly using a Riemann Sum

Writing this as

$$\lim_{n \to \infty}S_{nn} = \lim_{n \to \infty}\frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k}{n}+\frac{k}{n}\frac{1}{n^2}} $$

a technique that often works is to evaluate the double limit

$$\lim_{m \to \infty} \lim_{n \to \infty} S_{mn} = \lim_{m \to \infty} \lim_{n \to \infty} \frac{1}{n}\sum _{k=1}^n\frac{1}{1+\frac{k}{n}+\frac{k}{n}\frac{1}{m^2}} = \lim_{m \to \infty}\int_0^1 \frac{dx}{1 +x + x/m^2} = \int_0^1 \frac{dx}{1 +x} $$

where the last step is justified by DCT.

We can justify $ \lim_{n \to \infty} S_{nn} = \lim_{m \to \infty} \lim_{n \to \infty} S_{mn} = \log 2$ by showing one of the iterated limits exhibits uniform convergence.

An example with more details is given here. The argument for justifying the use of the double limit in this case will be similar.

It is easy to see that

$$ \begin{align} \left |\sum_{k=1}^n \frac{1}{n+k+k/n^2}-\sum_{k=1}^n \frac1{n+k}\right|&=\frac1{n^2}\sum_{k=1}^n \frac{k}{(n+k+k/n^2)(n+k)}\\\\ &\le \frac1{n^2}\sum_{k=1}^n \frac1k\tag 1 \end{align}$$

Then, using $\sum_{k=1}^n\frac1k =\gamma+\log(n) +O\left(\frac1n\right)$, we see that the limit of the left-hand side of $(1)$ as $n\to \infty$ is $0$.

Can you finish now?

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim _{n \to \infty }\sum _{k = 1}^{n}{1 \over n + k + k/n^{2}}} = \lim _{n \to \infty }\bracks{{n^{2} \over n^{2} + 1} \sum _{k = 1}^{n}{1 \over k + n^{3}/\pars{n^{2} + 1}}} \\[5mm] = &\ \lim _{n \to \infty }\braces{{n^{2} \over n^{2} + 1} \sum _{k = 1}^{\infty}\bracks{{1 \over k + n^{3}/\pars{n^{2} + 1}} - {1 \over k + n^{3}/\pars{n^{2} + 1} + n}}} \\[5mm] = &\ \lim _{n \to \infty }\braces{{n^{2} \over n^{2} + 1} \bracks{H_{\large n^{3}/\pars{n^{2} + 1} + n}\ -\ H_{\large n^{3}/\pars{n^{2} + 1}}}} \end{align} where $\ds{H_{z}}$ is a Harmonic Number.

In using the $\ds{H_{z}}$ asymptotic behavior as $\ds{\verts{z} \to \infty}$, it's straightforward found:

$$ \bbx{\lim _{n \to \infty }\sum _{k = 1}^{n}{1 \over n + k + k/n^{2}} = \ln\pars{2}} $$