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 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.  
