<s>[INST] 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 ?

 [/INST] 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. 
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