Evaluate the following limit Given positive integer $k$, evaluate
$$\lim_{n\to\infty}n\cdot\left(\sum_{i = 1}^n\left(\dfrac{i}{n}\right)^{k}\right)^{-1}$$
 A: $$
\begin{align*}
\lim_{n\to\infty}n\cdot\left(\sum_{i = 1}^n\left(\dfrac{i}{n}\right)^{k}\right)^{-1} &= 
\lim_{n\to\infty}\left(\frac1n \sum_{i = 1}^n\left(\dfrac{i}{n}\right)^{k}\right)^{-1}\\
\\
&=\left(\lim_{n\to\infty}\frac1n \sum_{i = 1}^n\left(\dfrac{i}{n}\right)^{k}\right)^{-1}\\
&= \left(\int_0^1 x^k dx\right)^{-1} \\ 
&= \left(\frac1{k+1}\right)^{-1} \\
&= k+1\\
\end{align*}
$$
A: The tricky part here is
$$\sum_{i=0}^n i^k = \frac{n^{k+1}}{k+1} + o(n^k)$$
Once you've got that 
$$\lim_{n\to\infty} n \left( \sum_{i=0}^n \left(\frac{i}{n}\right)^k \right)^{-1}  = \lim_{n\to\infty} n . n^{k} \left( \frac{n^{k+1}}{k+1} + \ldots \right)^{-1} = k+1$$
NOTE: 
To show the top result I used
$$ S_n = S_{n-1} + n^k $$ and assumed that $S_n = T_n + \alpha n^{k+1}$.
In that case 
$$ T_n = T_{n-1} + \alpha n^{k+1} - \alpha (n-1)^{k+1} + n^k $$
And we get a reduction in the order iff the first two terms ($n^{k+1}$ and 
$n^k$) cancel, which will only occur if $\alpha = 1/(k+1)$.
However I'm sure there's a more elegant way to do that.
