By considering:
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^1}{n^{2}} = \frac 1 2$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^2}{n^{3}} = \frac 1 3$$ $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^3}{n^{4}} = \frac 1 4$$
Determine if this is true: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}} = \frac 1 {{m+1}}$$
If it is, prove it.
If it is not, evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{{m+1}}$.
Since this is tagged algebra-precalculus, I will refrain from using the Euler-Maclaurin Sum Formula (which is exact in the case of polynomials).
Let's first show that $$ \sum_{k=1}^nk^m\text{ is a polynomial of degree }m+1\text{ in }n\text{ whose lead coefficient is }\frac{1}{m+1}\tag{1} $$
Since $\displaystyle\sum_{k=1}^n1=n$, we have shown that $(1)$ is true for $m=0$.
Suppose that $(1)$ is true for all $m<M$, then $$ k^{M+1}-(k-1)^{M+1}=\sum_{m=0}^{M}\binom{M+1}{m}(-1)^{M-m}k^m\tag{2} $$ If we sum $(2)$ for $k$ from $1$ to $n$, the telescoping sum on the left yields $$ n^{M+1}=\sum_{m=0}^{M}\binom{M+1}{m}(-1)^{M-m}\left(\sum_{k=1}^nk^m\right)\tag{3} $$ Isolating the $m=M$ term from the right-hand side of $(3)$ gives us $$ \sum_{k=1}^nk^M=\frac{1}{M+1}n^{M+1}-\frac{1}{M+1}\sum_{m=0}^{M-1}\binom{M+1}{m}(-1)^{M-m}\left(\sum_{k=1}^nk^m\right)\tag{4} $$ Since $(1)$ holds for each $m<M$, each $\displaystyle\left(\sum_{k=1}^nk^m\right)$ on the right-hand side of $(4)$ is a polynomial in $n$ of degree $m+1<M+1$. Thus, because of the $n^{M+1}$, the right-hand side of $(4)$ is a polynomial in $n$ of degree $M+1$ whose lead coefficient is $\dfrac{1}{M+1}$.
Thus, we have shown $(1)$ for all $m\ge0$.
Consider $$ \frac{1}{n^{m+1}}\sum_{k=1}^nk^m\tag{5} $$ The sum in $(5)$ is a polynomial in $n$ of degree $m+1$ whose lead coefficient is $\dfrac{1}{m+1}$. Therefore, $(5)$ becomes $$ \frac{1}{m+1}+O\left(\frac1n\right)\tag{6} $$ Thus, we get that $$ \lim_{n\to\infty}\;\frac{1}{n^{m+1}}\sum_{k=1}^nk^m=\frac{1}{m+1}\tag{7} $$
If you are familiar with it, this is a standard application of Stolz-Cesàro.
SC says that this limit is the same as
$$\lim_n \frac{(n+1)^m}{(n+1)^{m+1}-n^{m+1}}$$
It is an easy exercise (I'll leave this part to you) to prove that this limit is exactly $\frac{1}{m+1}$.
P.S. Stolz-Cesàro is for sequences exactly what l'Hôpital is for functions.
This follows easily from a comparison with an integral.
Since $$\int_0^{n} k^m dk \leq \sum_{k=1}^n k^m \leq \int_1^{n+1} k^m dk,$$ we have $$\frac{n^{m+1}}{m+1}\leq \sum_{k=1}^n k^m \leq \frac{(n+1)^{m+1}}{m+1}-\frac{1}{m+1}.$$ Divide by $n^{m+1}$ to get $$\frac{1}{m+1} \leq \frac{1}{n^{m+1}} \sum_{k=1}^n k^m \leq \frac{1}{m+1} \cdot \frac{(n+1)^{m+1}}{n^{m+1}} - \frac{1}{m+1} \cdot \frac{1}{n^{m+1}}.$$ Taking $n \rightarrow \infty$ gives the desired result.
What is true is that $$\sum_{k=1}^n k^m = {n^{m + 1}\over m+1} + P_m(n),$$ where $\deg(P) <= m.$ Dividing you will get $${1\over n^{m+1}}\sum_{k=1}^n k^m = {1\over m+1} + {1\over n^{m+1}}P_m(n) = {1\over m+1} + O(1/n). $$
By induction, using binomial theorem and telescoping, we have that
$$(k+1)^{m+1}-k^{m+1} = (m+1)k^{m} + O(k^{m-1}) $$$$\implies \sum_{k=1}^nk^m =\frac{(n+1)^{m+1}}{m+1}+O(n^m)=\frac{n^{m+1}}{m+1}+O(n^m)$$
and the resul follows.
