Evaluate limit of sums Evaluate $$\lim_{n \to \infty} \sum_{k=1}^n \frac{1}{n + \sqrt{k^2+n}}$$
I saw this as a Riemann Sum and tried to rewrite as
$$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(\frac{k}{n})^2 + \frac{1}{n}}}$$
but I can't say this is equal to $$\int_0^1 \frac{1}{1 + \sqrt{x^2 + 1}}$$ because there would be a too big error.
The answer should be $\ln 2$.
 A: A general approach for such "almost" Riemann sums is to evaluate as a double limit which, in this case, can be justified by uniform convergence of the inner limit:
$$\begin{align} \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/n}}  &= \lim_{n \to \infty} \lim_{m \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/m}}\\ &=   \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + (k/n)} \\ &= \int_0^1 \frac{dx}{1 +x } \\ &= \log 2 \end{align}$$
The relevant theorem is if $x_{nm} \to y_n$ uniformly as $m \to \infty$ and $x_{nm}$ converges as $n \to \infty$, then the double limit exists and 
$$\lim_{n \to \infty} x_{nn} = \lim_{n \to \infty} \lim_{m \to \infty}x_{nm} = \lim_{m \to \infty} \lim_{n \to \infty}x_{nm}$$
To show uniform convergence in this case, note that
$$\begin{align}\left|\frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/m}}  - \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n}   \right| &\leqslant  \frac{1}{n} \sum_{k=1}^n \frac{|1 + k/n - (1 + \sqrt{(k/n)^2 + 1/m})|}{|1+ k/n||1 + \sqrt{(k/n)^2 + 1/m}|} \\ &\leqslant  \frac{1}{n} \sum_{k=1}^n (\sqrt{(k/n)^2 + 1/m}-k/n) \\  &=\frac{1}{n} \sum_{k=1}^n  \frac{1/m}{\sqrt{(k/n)^2 + 1/m} + k/n}\\ &\leqslant \frac{1}{n} \sum_{k=1}^n \frac{1/m}{1/\sqrt{m}} \\ &= \frac{1}{\sqrt{m}}\end{align},$$
showing that
$$\lim_{m \to \infty} \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + \sqrt{(k/n)^2 + 1/m}} = \frac{1}{n} \sum_{k=1}^n \frac{1}{1 + k/n} $$
uniformly for all $n$.
Another approach would be to set bounds and apply the squeeze theorem.  
A: Squeezing:
$$\frac1{1+\sqrt{\left(\frac kn\right)^2+\frac1n}}\stackrel{(0)}<\frac1{1+\sqrt{\left(\frac kn\right)^2+0}}=\frac1{1+\frac kn}$$
And it is obvious that:
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac1{1+\frac kn}=\int_0^1\frac1{1+x}\ dx=\ln(2)$$
We then notice that:
$$\frac1{1+\sqrt{\left(\frac kn\right)^2+\frac1n}}\stackrel{(1)}>\frac1{1+\frac kn+\frac1{2k}}\stackrel{(2)}>\frac{1-\frac1{2k}}{1+\frac kn}\stackrel{(3)}>\frac1{1+\frac kn}-\frac1{2k}$$
And finally notice that:
$$\lim_{n\to\infty}\frac1n\sum_{k=1}^n\frac1{1+\frac kn}-\frac1{2k}\ge\ln(2)$$
which follows easily since
$$\frac1n\sum_{k=2}^n\frac1k<\frac1n\int_1^n\frac1x\ dx=\frac{\ln n}n$$
and this limit may be taken care of by L'Hospital's rule.

$(0)$ is motivated by the inequality $\frac1a<\frac1b$ if $0<b<a$.
$(1)$ is motivated by taking the line tangent to $\sqrt{a+x}$ at $x=0$.
$(2)$ is motivated by taking the line tangent to $\frac1{1+a+x}$ at $x=0$.
$(3)$ is just some basic algebra.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\begin{align}
\lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over n + \root{k^{2} + n}} & =
\lim_{n \to \infty}\sum_{k = 1}^{n}\pars{%
{1 \over n + \root{k^{2} + n}} - {1 \over k + n} + {1 \over k + n}}
\\[5mm] & =
\lim_{n \to \infty}\sum_{k = 0}^{n}\bracks{%
{k -  \root{k^{2} + n} \over \pars{n + \root{k^{2} + n}}\pars{k + n}} +
{1 \over k + n}}
\\[5mm] & =
\lim_{n \to \infty}\sum_{k = 0}^{n}\bracks{%
{1 \over k + n} -
{n \over \pars{n + \root{k^{2} + n}}\pars{k + \root{k^{2} + n}}\pars{k + n}}}
\end{align}

Note that

\begin{align}
&0 < \verts{\sum_{k = 0}^{n}
{n \over \pars{n + \root{k^{2} + n}}\pars{k + \root{k^{2} + n}}\pars{k + n}}} <
n\,{n \over \pars{n + \root{1 + n}}\root{n}n}
\\[5mm] \stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,&\ {1 \over \root{n}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\to}\,\,\,{\large 0}
\end{align}

Then,

\begin{align}
\lim_{n \to \infty}\sum_{k = 1}^{n}{1 \over n + \root{k^{2} + n}} & =
\lim_{n \to \infty}\sum_{k = 0}^{n}{1 \over k + n} =
\lim_{n \to \infty}\pars{{1 \over n}\sum_{k = 0}^{n}{1 \over 1 + k/n}} =
\int_{0}^{1}{\dd x \over 1 + x} =\
\bbox[10px,#ffe,border:1px dotted navy]{\ds{\ln\pars{2}}}
\end{align}
