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  ,
 A: Also, you can use Stolz: https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem
$$\lim_{n\rightarrow+\infty}\frac{1^4+2^4+...+n^4}{n^5}=\lim_{n\rightarrow+\infty}\frac{n^4}{n^5-(n-1)^5}=\frac{1}{5}.$$
A: You can’t do this step
$$\lim_{n\rightarrow \infty } \frac{ \frac{1^4}{n^4} + \frac{2^4}{n^4} +...+1 }{n}\color{red}{=\frac{0 + 0 +0 +...+1 }{n}= 0} $$
since you have infinitely many terms which tend to 0 and their sum not necessarily is $0$.
Think to $\sum_1^{\infty} \frac1k$ which diverges.
A: You were going right till the very last step. You need to apply the limit after computing the sum. The limit converts to Riemann integral:
$$\lim_{n\to \infty}\sum_{k=1}^{n} \frac{k^4}{n^5} = \int_{0}^{1} x^4 dx = \frac{1}{5}$$

Since you are not familiar with integration, another method is to find $\sum_{k=1}^{n} k^4$ explicitly. Only the leading coefficient, ie the coefficient of $n^5$ needs to be found.
$$\sum_{k=1}^nk^4=\frac1{30}n(n+1)(2n+1)(3n^2+3n-1)$$
So we can write $\sum_{k=1}^nk^4 = \frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} -\frac{n}{30}$. Thus if you compute the limit
$$\lim_{n\to \infty}\sum_{k=1}^n\frac{k^4}{n^5} = \lim_{n\to\infty} \frac{\frac{n^5}{5} + \frac{n^4}{2} + \frac{n^3}{3} -\frac{n}{30}}{n^5}\\
\lim_{n\to\infty} \frac{\frac{1}{5} + \frac{1}{2n} + \frac{1}{3n^2} -\frac{1}{30n^4}}{1} = \frac{1}{5}$$
A: 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.
A: The problem is that while each term tends to zero, the number of terms tends to infinity at the same time.
The rule “the limit of a sum is the sum of the limits” only applies if you have a finite sum with a fixed number of terms (each of which has a finite limit).
(Also, you can't really write
$$\lim_{n \to \infty} (\dots) = (\text{some expression involving $n$})$$
since there can be no $n$ left in your expression after you have let $n \to \infty$. The symbol $n$ stands for the same number everywhere, so you can't selectively let some $n$ tend to $\infty$ while leaving other $n$ unaffected.)
