Limit $ \lim_{n \to \infty} \sum_{k=1}^n\frac 1 { \sqrt{n^2+k} } $ (1)
$$ \lim_{n \to \infty} \sum_{k=1}^n \frac 1 { \sqrt{n^2+k} }    $$
(2)
$$ \lim_{n\to\infty} \frac {1+\sqrt[n]2 + \sqrt[n]3 + ... \sqrt[n]n} {n}  $$
The answers should both be 1.. any hints?
 A: For the first problem, note that if $1\le k\le n$, then 
$$n^2\lt n^2+k\le n^2+n\lt\left(n+\frac{1}{2}\right)^2.$$
Thus 
$$\frac{1}{n+\frac{1}{2}}\le \frac{1}{\sqrt{n^2+k}}\lt \frac{1}{n}.$$
Our sum is therefore between $\frac{n}{n+\frac{1}{2}}$ and $1$. 
Squeeze.
A: Here is an alternative proof for (2) that is not in the suggested answers (without employing Cezaro-Stolz).
$$\lim_{n \rightarrow \infty} \frac{1+(2)^{1/n} + ... + (n)^{1/n}}{n} \\ 
= \lim_{n \rightarrow \infty} \frac{n^{1/n}\cdot ((1/n)^{1/n}+(2/n)^{1/n} + ... + (n/n)^{1/n})}{n} \\
= \lim_{n \rightarrow \infty} \frac{((1/n)^{1/n}+(2/n)^{1/n} + ... + (n/n)^{1/n})}{n} \\
= \lim_{m,n \rightarrow \infty} \frac{((1/n)^{1/m}+(2/n)^{1/m} + ... + (n/n)^{1/m})}{n} \\
= \lim_{m \rightarrow \infty} \int_{0}^{1} x^{1/m}\; dx \\
= \lim \frac{1^{1+1/m}}{1+1/m} \\
= 1$$
NOTE: that $\lim n^{1/n} = \exp(\lim \frac{\log(n)}{n}) = \exp(0) = 1$ and so the second step is allowed. Also, you probably want to tighten up the analysis regarding uniform convergence though.
