Proof equation $\lim_{n \to \infty}^{}\sum_{k=1}^{n}k^{p}/(n+1)^{p}= \frac{1}{p+1}$

Question

Just tried some approach (Stolz-Cesàro theorem and sandwich theorem) can't prove this equation $$\lim_{n \to \infty}^{}\frac{\sum\limits_{k=1}^{n}k^{p}}{(n+1)^{p}} =\frac{1}{p+1}$$

Answer 1

I guess your question is wrong

$$ \lim_{n\to{\infty}}\frac{\sum_{k=1}^{n}k^p}{(n+1)^{p+1}}=\frac{1}{p+1} $$

Answer 2

[Assuming the correction in Eeyore Ho's answer is correct]

The easiest way must be by using the bounds: $$\int_1^n x^p\; dx\leq \sum_{k=1}^n k^p\leq \int_1^{n+1} x^p\; dx$$

However, you can also avoid using calculus. Instead it can be proven by induction.

We know that each sum of $p$th powers up till $n$ equals some polynomial of degree $p+1$ in $n$ (you don't need to assume this for the following argument). This is clear for $p=0$. Now, we can find a recurrence relation for the coefficients in general.

Let $m$ be a positive integer. Suppose we have found an expression $\sum_{k=1}^n k^p=\sum_{j=0}^{p+1} (a_{j,p}k^j)$ for all $p<m$ where each the list $(a_{j,p})_j$ is the set of associated constant coefficients.

We rearrange the sum of $m$th powers as follows $$\sum_{k=1}^n k^m=\sum_{k=1}^n k\cdot k^{m-1}=\sum_{k=1}^n\sum_{j=1}^k k^{m-1}=\sum_{l=1}^n\sum_{q=l}^n q^{m-1}=\sum_{l=1}^n\left(\sum_{q=1}^nq^{m-1}-\sum_{q=1}^{l-1}q^{m-1}\right)$$$$=\sum_{l=1}^{n+1}\left(\sum_{q=1}^nq^{m-1}-\sum_{q=1}^{l-1}q^{m-1}\right)=(n+1)\sum_{q=1}^nq^{m-1}-\sum_{l=1}^{n+1}\sum_{q=1}^{l-1}q^{m-1}$$$$=(n+1)\sum_{q=1}^nq^{m-1}-\sum_{r=1}^n\sum_{q=1}^rq^{m-1}$$

By the induction hypothesis, then $$\sum_{k=1}^n k^m=(n+1)\sum_{b=0}^m(a_{b,m-1}n^b)-\sum_{r=1}^n\sum_{c=0}^m (a_{c,m-1}r^c)$$

Further evaluating, we get $$=(n+1)\sum_{b=0}^m(a_{b,m-1}n^b)-\sum_{c=0}^m \left(a_{c,m-1}\left(\sum_{r=1}^n r^c\right)\right)$$$$=(n+1)\sum_{b=0}^m(a_{b,m-1}n^b)-\sum_{c=0}^m \left(a_{c,m-1}\left(\sum_{s=0}^{c+1}a_{d,c}n^d\right)\right)$$$$=(n+1)\sum_{b=0}^m(a_{b,m-1}n^b)-\sum_{c=0}^m\sum_{s=0}^{c+1}(a_{c,m-1}a_{d,c}n^d)$$

This affirms (by strong induction) that the general sum of $p$th powers up till the number $n$ is a polynomial in $n$ of degree $p+1$.

Now assume also that we have proven for the case of $p=m-1$ that the term $a_{p+1,p}=a_{m,m-1}$ (i.e. the coefficient of $n^m$ in the polynomial expansion of the sum of $(m-1)$th powers up till $n$) is equal to $\frac{1}{m}$ (it clearly holds for $m-1=0$).

Then, picking the coefficient of the $n^{m+1}$ in our final summation expression, we have $$a_{m+1,m}=a_{m,m-1}-a_{m,m-1}a_{m+1,m}$$

But, by assumption, we already have $a_{m,m-1}=\frac{1}{m}$. So rearranging, we get $a_{m+1,m}=\frac{1}{m+1}$.

So the coefficient of the $(p+1)$th power of $n$ appearing in $\sum_{k=1}^n k^p$ is $\frac{1}{p+1}$ for all nonnegative $p$. The desired result in your question then immediately follows.

(feel free to comment or edit for any corrections)