Evaluate $\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$ 
Evaluate
  $\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$ using Stolz-Cesaro theorem . 
  Now this is my attempt :$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p+(2n)^p-1^p-3^p-\dots-(2n-1)^p}{(n+1)^{p+1}-n^{p+1}}$=$\lim\limits_{n\to \infty}\frac{(2n)^p}{(n+1)^{p+1}-n^{p+1}}$ (and now i was thinking to use the binomial theorem ) 
  $\lim\limits_{n\to \infty}\frac{(2n)^p}{n^{p+1}+{p+1 \choose p}n^p+\dots+1-n^{p+1}}$,which will eventually lead to  the answer : $\frac{2}{p+1}$ . Is this correct ?

 A: Let $ n $ be a positive integer greater than $ 1 \cdot $
Since $ \sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}=\sum\limits_{k=1}^{2n}{k^{p}}-\sum\limits_{k=1}^{n}{\left(2k\right)^{p}}=\sum\limits_{k=1}^{n}{k^{p}}+\sum\limits_{k=n+1}^{2n}{k^{p}}-2^{p}\sum\limits_{k=1}^{n}{k^{p}}$ $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left(1-2^{p}\right)\sum\limits_{k=1}^{n}{k^{p}}+\sum\limits_{k=1}^{n}{\left(n+k\right)^{p}} \cdot $
We have : $ \frac{1}{n^{p+1}}\sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}=\left(1-2^{p}\right)\left(\frac{1}{n}\sum\limits_{k=1}^{n}{\left(\frac{k}{n}\right)^{p}}\right)+\frac{1}{n}\sum\limits_{k=1}^{n}{\left(1+\frac{k}{n}\right)^{p}}$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underset{n\to +\infty}{\longrightarrow}\left(1-2^{p}\right)\int\limits_{0}^{1}{x^{p}\,\mathrm{d}x}+\int\limits_{0}^{1}{\left(1+x\right)^{p}\,\mathrm{d}x} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underset{n\to +\infty}{\longrightarrow}\frac{1-2^{p}}{p+1}+\frac{2^{p+1}-1}{p+1} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{n^{p+1}}\sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}\underset{n\to +\infty}{\longrightarrow}\frac{2^{p}}{p+1} $
A: No, not correct. 
\begin{align*}
&\quad \ \ \frac {(2n)^p} {(n+1)^{p+1} - n^{p+1}} \\
&=\frac 1 n \cdot \frac {2^p} {(1 + 1/n)^{p+1} - 1} \\
&\sim \frac 1 n \cdot \frac {2^p} {(p+1)/n} \\
&= \frac {2^p} {p+1}.  \tag {$n \to \infty$}
\end{align*}
$p$ is a real positive number, so generally the binomial theorem fails. Instead, a frequently used equivalent infinitesimal would work. 
