Proof equation: $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n(k(k+1)...(k+m-1))}{\sum_{k=1}^nk^m}=1$ I need help to proof next equation: 
$\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n(k(k+1)...(k+m-1))}{\sum_{k=1}^nk^m} = 1$
I know, that 
$\lim\limits_{n\to\infty}\frac{\sum_{k=1}^nk^m}{{n^{m+1}}} = {\frac{1}{m+1}}$
so
$\frac{\sum_{k=1}^n(k(k+1)...(k+m-1))}{n^{m+1}}$ should equals to (m+1)

Stolz–Cesàro theorem
$\lim\limits_{n\to\infty}\frac{x_n}{y_n} = \lim\limits_{n\to\infty}\frac{x_n-x_{n-1}}{y_n-y_{n-1}}$
$\lim\limits_{n\to\infty}\frac{n(n+1)...(n+m-1)}{n^m} = \lim\limits_{n\to\infty}\frac{n^m (...)}{n^m} = \lim\limits_{n\to\infty}{(1+1/n)(1+2/n)...(1+(m-1)/n) = 1}$
Awesome! thanks!
 A: Note that
$$\sum_{k=1}^n k(k+1)...(k+m-1)=m!\sum_{k=1}^n\binom{k+m-1}{m}$$
is a well known binomial sum. 
Or use $m+1=(m+k)-(k-1)$ to arrive at a telescoping sum.

And as noted in the comment, the Cesaro-Stolz theorem gives the limit without computing any sums.
A: Let $\mathcal{N}_n$ and $\mathcal{D}_n$ be the numerator and denominator in the limit.
For the numerator, we have
$$\begin{align}
\mathcal{N}_n 
= \sum_{k=1}^n \prod_{\ell=0}^{m-1}(k+\ell)
=&\frac{1}{m+1}\sum_{k=1}^n((k+m)-(k-1))\prod_{\ell=0}^{m-1}(k+\ell)\\
=& \frac{1}{m+1}\sum_{k=1}^n \left(\prod_{\ell=0}^{m}(k+\ell) - \prod_{\ell=0}^{m}(k-1+\ell)\right)\\=& \frac{1}{m+1}\prod_{\ell=0}^m(n+\ell)
\end{align}
$$
This implies
$$\lim_{n\to\infty} \frac{1}{n^{m+1}} \mathcal{N}_n = \frac{1}{m+1}\lim_{n\to\infty}\prod_{\ell=0}^m\left(1 + \frac{\ell}{n}\right) = \frac{1}{m+1}$$
For the denominator,
$$\frac{1}{n^{m+1}}\mathcal{D}_n = \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^m$$
This has the form of a Riemann sum. As a result,
$$\lim_{n\to\infty} \frac{1}{n^{m+1}}\mathcal{D}_n = \int_0^1 x^m dx = \frac{1}{m+1}$$
Combine these two results, we get
$$\lim_{n\to\infty}\frac{\mathcal{N}_n}{\mathcal{D}_n} =
\lim_{n\to\infty}\frac{\frac{1}{n^{m+1}}\mathcal{N}_n}{\frac{1}{n^{m+1}}\mathcal{D}_n} = \frac{\frac{1}{m+1}}{\frac{1}{m+1}} = 1$$
