Finding limits using definite integrals $\lim_{n\to\infty}\sum^n_{k=1}\frac{k^{4}}{n^{5}}$ 
Find the limit of $\displaystyle\lim_{n\to\infty}\frac 1 {n^5}(1^4+2^4...+n^4)$ using definite integrals.

It's equal to: $\displaystyle\lim_{n\to\infty} \sum^n_{i=1}\frac 1 i$ but now I'm not sure how to turn it to an integral.
$\Delta x_i=\frac 1 n, f(x_i)=1$ so the integral would be: $\displaystyle\int 1dx$ ? How can I find the bounds?
 A: Hint: 
Summation of the series using definite integral:
$$\lim \limits_{n\to \infty }\frac{1}{n} \sum \limits^{h(n)}_{r=g(n)}f(\frac{r}{n})=\int \limits^{b}_{a}f(x)dx$$
Where 
1.$$\sum \to \int$$
2.$$\frac{r}{n} \to x$$
3.$$\frac{1}{n} \to dx$$
4.$$a=\lim \limits_{n\to \infty }\frac{g(n)}{n}$$
5.$$b=\lim \limits_{n\to \infty }\frac{h(n)}{n}$$
A: Draw the curve $y=x^4$. By area comparison, we have 
$$\int_0^n x^4\,dx\lt 1^4+2^4+\cdots+n^4\lt \int_0^{n+1} x^4\,dx.$$
It follows that
$$\frac{1}{5}\lt \frac{1^4+2^4+\cdots +n^4}{n^5}\lt \frac{1}{5}\cdot \frac{(n+1)^5}{n^5}.$$
Since $\lim_{n\to\infty}\frac{(n+1)^5}{n^5}=1$, it follows by Squeezing that our limit is $\frac{1}{5}$.
A: That can be tackled also without integrals. Obviously:
$$ 4!\binom{n}{4}\leq n^4 \leq 4!\binom{n+3}{4} \tag{1}$$
and by summing both sides over $n=1,2,\ldots,N$ through a well-known combinatorial identity we get:
$$ 4!\binom{N+1}{5}\leq\sum_{n=1}^{N}n^4\leq 4!\binom{N+4}{5}\tag{2} $$
and by dividing both sides by $N^5$ and letting $N\to +\infty$:
$$ \lim_{N\to +\infty}\frac{1}{N^5}\sum_{n=1}^{+\infty}n^4 = \frac{4!}{5!}=\color{red}{\frac{1}{5}}.\tag{3}$$
A: Although not asked by the OP, I thought it might be interesting to mention that the sum $\sum_{k=1}^n k^4$ is well known (http://mathworld.wolfram.com/PowerSum.html) and can be written 
$$\sum_{k=1}^n k^4 =\frac{1}{5}n^5 +\frac12  n^4+\frac13 n^3-\frac{1}{30}n$$
Dividing by $n^5$ and letting $n\to \infty$ recovers the expected result
$$\lim_{n\to \infty}\frac{1}{n^5}\sum_{k=1}^n k^4 =\frac{1}{5}$$
A: Hint: rewrite the sum as:
$$\frac1n\sum_{k=1}^n\Bigl(\frac kn\Bigr)^4.$$
This is an upper Riemann sum for the function $x^4$ on the interval $[0,1]$.
A: More generally,
and with no originality,
let $f$ be a function with
$f'(x) \ge 0$
and $f(x) > 0$
Then
$f(n)
\le \int_n^{n+1} f(x) dx
\le f(n+1)
$.
(If
$f'(x) \le 0$
and $f(x) > 0$,
then
the inequalities are reversed
so
$f(n)
\ge \int_n^{n+1} f(x) dx
\ge f(n+1)
$.)
Summing for
$n = 1$ to $N-1$,
$\sum_{n=1}^{N-1}f(n)
\le \int_1^{N} f(x) dx
\le \sum_{n=1}^{N-1} f(n+1)
$
or
$ f(0)
\le \sum_{n=1}^{N}f(n)-\int_1^{N} f(x) dx
\le f(N)
$.
Therefore,
if
$\frac{f(N)}{\int_1^{N} f(x) dx}
\to 0
$,
since 
$\frac{\int_0^{1} f(x) dx}{\int_1^{N} f(x) dx}
\to 0
$,
$\frac1{\int_0^{N} f(x) dx}\sum_{n=1}^{N}f(n)
\to 1
$.
Letting
$f(x)
=x^p
$
with $p \ge 1$,
$\frac{p+1}{N^{p+1}}\sum_{n=1}^{N}n^p
\to 1
$
or
$\frac1{N}\sum_{n=1}^{N}\left(\frac{n}{N}\right)^p
\to \frac1{p+1}
$.
A: Another solution that does not require integrals: using Cesaro-Stolz,
$$
\lim_{n\to\infty}\frac{(1^4+2^4+\dots+n^4)}{n^5} =
\lim_{n\to\infty}\frac{(n+1)^4}{(n+1)^5-n^5} = \cdots
$$
