Finding $\lim_{n\to\infty}\frac{1^p+3^p+...+(2n+1)^p}{n^{p+1}}$ I'm trying to solve the following problem:
Find $$\lim_{n\to\infty}\frac{1^p+3^p+\ldots+(2n+1)^p}{n^{p+1}}$$
What I've got so far:
My idea is to use Stolz-Cesaro theorem, which implies that:
$$
\lim_{n\to\infty}\frac{1^p+3^p+\ldots+(2n+1)^p}{n^{p+1}}=\lim_{n\to\infty}\frac{(2n+3)^p}{(n+1)^{p+1}-n^{p+1}}
$$
From there:
$$\frac{(2n+3)^p}{(n+1)^{p+1}-n^{p+1}}=\frac{(2n+3)^p}{(2n+1)^{\frac{p+1}{2}}}$$
So, if my calculations are correct I only have to find the limit of the term in the RHS on the last line, that's unless the approach is somewhere entirely different.
Thanks in advance!
 A: By Riemann sums, for any $p>-1$:
$$ \frac{1}{n}\sum_{k=0}^{n}\left(\frac{2k+1}{n}\right)^p \xrightarrow{n\to +\infty}\int_{0}^{1}(2x)^p\,dx = \color{red}{\frac{2^p}{p+1}}.$$
A: You are on the right track: as $n\to +\infty$,
$$\frac{(2n+3)^p}{(n+1)^{p+1}-n^{p+1}}=\frac{2^pn^p(1+\frac{3}{2n})^p}{n^{p+1}\left(\left(1+\frac{1}{n}\right)^{p+1}-1\right)}=\frac{2^p(1+o(1))}{n\left(1+\frac{p+1}{n}+o(1/n)-1\right)}\to {\frac{2^p}{p+1}}.$$
A: HINT:
$$\dfrac1n\sum_{r=0}^n\dfrac{(2r+1)^p}{n^p}=\dfrac{(2n+1)^p}{n^{p+1}}+\dfrac1n\sum_{r=1}^n\dfrac{(2r-1)^p}{n^p}$$
Now $$\dfrac1n\sum_{r=1}^n\dfrac{(2r-1)^p}{n^p}=\underbrace{\dfrac1n\sum_{r=1}^{2n}\left(\dfrac rn\right)^p}_{(1)}-\underbrace{\dfrac1n\sum_{r=1}^n\left(\dfrac{2r}n\right)^p}_{(2)}$$
For $(1)$ see See also : Find $\lim\limits_{n \to \infty} \frac{1}{n}\sum\limits^{2n}_{r =1} \frac{r}{\sqrt{n^2+r^2}}$
Now for $(2)$ use $$\lim_{n \to \infty} \frac1n\sum_{r=1}^n f\left(\frac rn\right)=\int_0^1f(x)dx$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
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\begin{align}
&\lim_{n \to \infty}{1^{p} + 3^{p} + \cdots + \pars{2n + 1}^{\,p} \over
n^{p + 1}}
=
\lim_{n \to \infty}{\sum_{k = 0}^{n}\pars{2k + 1}^{\,p} \over n^{p + 1}}
\\[5mm] = &\
\lim_{n \to \infty}{\sum_{k = 0}^{n + 1}\pars{2k + 1}^{\,p} -
\sum_{k = 0}^{n}\pars{2k + 1}^{\,p} \over \pars{n + 1}^{p + 1} - n^{p + 1}}
\qquad\pars{~Stolz-Ces\grave{a}ro\ Theorem~}
\\[5mm] = &\
\lim_{n \to \infty}{\pars{2n + 3}^{\,p} \over n^{p + 1}
\bracks{\pars{1 + 1/n}^{p + 1} - 1}} =
2^{p}\lim_{n \to \infty}{\bracks{1 + 3/\pars{2n}}^{\,p} \over n
\bracks{\pars{1 + 1/n}^{p + 1} - 1}}\label{1}\tag{1}
\end{align}

Note that
  $\ds{{\bracks{1 + 3/\pars{2n}}^{\,p} \over n
\bracks{\pars{1 + 1/n}^{p + 1} - 1}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
{1 + 3p/\pars{2n} + 9p\pars{p - 1}/\pars{8n^{2}}\over
n\bracks{\pars{p + 1}/n + \pars{p + 1}p/\pars{2n^{2}}}}}$

\begin{align}
&\mbox{such that}\quad\lim_{n \to \infty}{\bracks{1 + 3/\pars{2n}}^{\,p} \over n
\bracks{\pars{1 + 1/n}^{p + 1} - 1}} = {1 \over p + 1}
\\[5mm] &\
\mbox{and}\quad\pars{~\mrm{see}\ \eqref{1}~}\quad
\bbx{\lim_{n \to \infty}{1^{p} + 3^{p} + \cdots + \pars{2n + 1}^{\,p} \over
n^{p + 1}}
=
{2^{p} \over p + 1}}
\end{align}
