limit without expansion I was solving this limit  instead of using sum of expansion i  did this and got zero  but answer is 1/5 by expansion why this is wrong
$$\lim_{n\rightarrow \infty } \frac{1^4 + 2^4 + \ldots +  n^4}{n^5}$$
$$\lim_{n\rightarrow \infty } \frac{n^4( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n^5}$$
$$= \lim_{n\rightarrow \infty } \frac{( \frac{1^4}{n^4} + \frac{2^4}{n^4} +\ldots + 1 )}{n}$$
$$= \frac{(0 + 0 +0 \ldots + 1 )}{n} = 0$$ this is how i got  ,

If
$m > 1$ then
$k^m
\lt \int_k^{k+1} x^m dx
\lt (k+1)^m
$.
Summing the left inequality
from $1$ to $n-1$,
$\begin{array}\\
\sum_{k=1}^{n-1}k^m
&\lt \sum_{k=1}^{n-1}\int_k^{k+1} x^m dx\\
&= \int_1^{n} x^m dx\\
&= \dfrac{x^{m+1}}{m+1}\big|_1^{n}\\
&< \dfrac{n^{m+1}-1}{m+1}+(n+1)^m\\
&< \dfrac{n^{m+1}}{m+1}\\
\text{so}\\
\sum_{k=1}^{n}k^m
&< \dfrac{n^{m+1}}{m+1}+n^m\\
\text{so}\\
\dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m
&< \dfrac{1}{m+1}+\dfrac1{n}\\
\end{array}
$
Similarly,
summing the right inequality
from $0$ to $n-1$,
$\begin{array}\\
\sum_{k=0}^{n-1}(k+1)^m
&\gt \sum_{k=0}^{n-1}\int_k^{k+1} x^m dx\\
&= \int_0^{n} x^m dx\\
&= \dfrac{x^{m+1}}{m+1}\big|_0^{n}\\
&= \dfrac{n^{m+1}}{m+1}\\
\text{so}\\
\sum_{k=1}^{n}k^m
&> \dfrac{n^{m+1}}{m+1}\\
\text{so}\\
\dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m
&> \dfrac{1}{m+1}\\
\end{array}
$
Therefore
$0
\lt \dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m
- \dfrac{1}{m+1}
\lt \dfrac1{n}
$
so
$\lim_{n \to \infty} \dfrac1{n^{m+1}}\sum_{k=1}^{n}k^m
=\dfrac1{m+1}$.
Note:
Nothing original here,
though I took pains
to make the result
come about
as painlessly as possible.
