Find $ \lim_{k \rightarrow \infty } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$ How to compute the limit 
$$ \lim_{k \rightarrow   \infty  } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}$$
Does exist an explicit formula for finding the numerator?
 A: As mentioned by Minus One-Twelfth, 
$$\lim_{k \rightarrow   \infty  } \frac{ 1^n + 2^n + \ldots + k^{n+1}}{k^{n+2}} = \dfrac {1}{n+2},$$
which is easily proven using Riemann's sums/Faulhaber's formula:
Your limit equals 
$$ \lim_{k \rightarrow   \infty  } \dfrac{ 1}{n+2}\times\frac{k^{n+2}}{k^{n+2}+k^{n+1}} = \dfrac{1}{n+2}$$
A: The required limit is $1/(n+2)$. There probably does not exist a closed formula for the numerator. However, the numerator can be estimated by use of an integral, as follows.
Note that
$$\frac{k^{n+2}}{n+2}=\int_0^{k}x^{n+1}\,\mathrm dx < \sum_{x=0}^kx^{n+1}<\int_1^{k+1}x^{n+1}\,\mathrm dx=\frac{(k+1)^{n+2}}{n+2}-\frac{1}{n+2}$$
Therefore,
$$\frac{k}{(n+2)(k+1)}<\frac{\sum_{x=0}^kx^{n+1}}{(k+1)k^{n+1}}<(1+\frac{1}{k})^{n+1}\frac{1}{n+2}-\frac{1}{(k+1)k^{n+1}(n+2)}$$
and both the r.h.s and the l.h.s tend to $\frac{1}{n+2}$ as $k\to\infty$.
A: By the Stolz-Cesro Theorem , one has
\begin{eqnarray}
&&\lim_{k \rightarrow   \infty  } \frac{ \sum_{x = 0}^{k} {x^{n + 1}}} {(k + 1)k^{n + 1}}\\
&=&\lim_{k \rightarrow   \infty  } \frac{{(k+1)^{n + 1}}} {(k + 2)(k+1)^{n + 1}-(k + 1)k^{n + 1}}\\
&=&\lim_{k \rightarrow   \infty  } \frac{1} {(k + 2)-(k + 1)\bigg(\frac{k}{k+1}\bigg)^{n + 1}}\\
&=&\lim_{k \rightarrow   \infty  } \frac{1} {(k + 2)-(k + 1)\bigg(1-\frac{1}{k+1}\bigg)^{n + 1}}\\
&=&\lim_{k \rightarrow   \infty  } \frac{1} {(k + 2)-(k + 1)\bigg(1-(n+1)\frac{1}{k+1}+O(\frac1{(k+1)^2})\bigg)}\\
&=&\frac 1{n+2}.
\end{eqnarray}
