<s>[INST] Help with $\lim \frac {{1}^{p} + 3^p + ... + (2n+1)^p}{n^{p+1}}$. I'm using Stolz–Cesàro theorem  to show that this limit  = $\frac {2^p}{P+1}$. I took $(X_{n})$ = $\sum_{i=0}^{n}{2i+1}^p$ and $(Y_n)$ = $n^{p+1}$. Then, by the theorem, $\lim = \frac {X_{n+1}-X_n}{Y_{n+1}-Y_n}$.
In the numerator I get $(2n+3)^{p}$. I've tried to expand this and I have leftover terms $$2 + \frac{6}{p} + \frac{9}{2p^{2}}+...$$ after canceling out. I don't know what to do with these, or if I did the expansion correctly.
 [/INST] Try to link the summation to a Riemann sum:
\begin{align}
& \frac{1^p + 3^p + \cdots + (2n + 1)^p}{(n + 1)^{p + 1}} \\
= & \frac{1}{2}\sum_{i = 1}^{n + 1}\left(\frac{2i - 1}{n + 1}\right)^p\times\frac{2}{n + 1} \\
\to & \frac{1}{2} \int_0^2 x^p dx = \frac{2^p}{p + 1},
\end{align}
during which we partitioned interval $[0, 2]$ by an equally-spaced partition 
$$\left\{0, \frac{1}{n + 1}, \frac{3}{n + 1}, \ldots, \frac{2n + 1}{n + 1}, 2\right\}.$$
Since $\lim_{n \to \infty} \frac{(n + 1)^p}{n^p} = 1$, the result follows. 
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