Prove that $\lim_{n\rightarrow\infty}\frac{1}{n^{p+1}}\sum_{k=1}^{n}k^{p}=\frac{1}{p+1} $

Question

I need to prove that $$ \lim_{n\to\infty}\frac{1}{n^{p+1}}\sum_{k=1}^n k^p = \frac{1}{p+1} $$

By Stulz lemma, $$\frac{(n+1)^{p}}{(n+1)^{p+1}-n^{p+1}}=\lim_{n\rightarrow\infty}\frac{1}{n^{p+1}}\sum_{k=1}^{n}k^{p}=\frac{1}{p+1}$$ and $$\frac{(n+1)^{p}}{(n+1)^{p+1}-n^{p+1}}\geq\frac{1}{\frac{p+1+\frac{1}{n}}{1+\frac{1}{n}}}\rightarrow \frac{1}{1+p}$$

And know I'm stuck..

Answer 1

Use the Faulhaber formula: $$\sum_{k=1}^n k^p = \dfrac1{p+1} \sum_{j=0}^p (-1)^j \dbinom{p+1}j B_j n^{p+1-j}$$ where $B_j$ are the Bernoulli numbers.

Answer 2

Hint:

$$\int_1^nx^p dx \leq \sum_{k=1}^n k^p \leq \int_1^{n+1}x^p dx$$

Precalculus:

Use induction. For the upper bound assume that

$$\sum_{k=1}^n k^p \leq \frac{(n+1)^{p+1}}{p+1}$$

holds. Then show that $(n+1)^p + \frac{(n+1)^{p+1}}{p+1} \leq \frac{(n+2)^{p+1}}{p+1}$ to conclude that $\sum_{k=1}^{n+1} k^p \leq \frac{(n+2)^{p+1}}{p+1}$ and since it holds for $n=1$ it follows by induction that the inequality hold for all $n$. For the lower bound assume that

$$\frac{n^{p+1}-1}{p+1}\leq \sum_{k=1}^n k^p$$

holds. Then show that $\frac{n^{p+1}-1}{p+1} + (n+1)^p\geq \frac{(n+1)^{p+1}}{p+1}$ and conclude.

Answer 3

We have: $$ k(k-1)\cdot\ldots\cdot(k-p+1)=p!\binom{k}{p}\leq k^p \leq p!\binom{k+p}{p}=(k+p)\cdot\ldots\cdot(k+1). $$ Summing over $k$: $$ p!\binom{n+1}{p+1}\leq\sum_{k=1}^{n}k^p\leq p!\binom{n+p+1}{p+1} $$ hence: $$ \lim_{n\to +\infty}\frac{1}{n^{p+1}}\sum_{k=1}^{n}k^p = \frac{1}{p+1}$$ follows by squeezing.

Answer 4

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,{\rm Li}_{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{n\ \to\ \infty}{1 \over n^{p + 1}}\sum_{k\ =\ 1}^{n}k^{p} ={1 \over p + 1}:\ {\large ?}}$.

\begin{align}&\color{#66f}{\large% \lim_{n\ \to\ \infty}{1 \over n^{p + 1}}\sum_{k\ =\ 1}^{n}k^{p}} =\lim_{n\ \to\ \infty}{1 \over n}\sum_{k\ =\ 1}^{n}\pars{k \over n}^{p} =\int_{0}^{1}x^{p}\,\dd x =\left.{x^{p + 1} \over p + 1}\right\vert_{\, x\ =\ 0}^{\,x\ =\ 1} =\color{#66f}{\large{1 \over p + 1}} \end{align}