TITLE: Find limits of sequences
QUESTION [7 upvotes]: Prove that 
$$a) \lim\limits_{n \to \infty} (\frac{1^p+2^p+...+n^p}{n^p} - \frac{n}{p+1})=\frac{1}{2},$$
$$b) \lim\limits_{n \to \infty} \frac{1^p+3^p+...+(2n-1)^p}{n^{p+1}}=\frac{2^p}{p+1},$$
where is $p \in \Bbb N $.
Thanks to Stolz–Cesàro theorem in $a)$ I went to $$\lim\limits_{n \to \infty} \frac{(n+1)^p}{(n+1)^p-n^p}$$ which after dividing by $n^p$ goes to
$$\lim\limits_{n \to \infty} \frac{1+\frac{p-1}{n}+\frac{p(p-1)}{2n^2}+...+\frac{1}{n^p}}{\frac{p-1}{n}+\frac{p(p-1)}{2n^2}+...+\frac{1}{n^p}} = \frac{1}{\infty}$$
At the same time $$\lim\limits_{n \to \infty} \frac{n}{p+1}=\infty,$$
so initial limit should be $\frac{1}{\infty}-\infty = ?$
I can't figure out what I'm missing ( $b)$ seems to complicated to me so I didn't even try it).

REPLY [5 votes]: We know that the sum of the $p^{th}$ powers of the integers is a polynomial of degree $p+1$ in $n$, let $S(n):=\alpha n^{p+1}+\beta n^p+R(n)$.
Then
$$n^p=S(n)-S(n-1)=\alpha n^{p+1}-\alpha (n-1)^{p+1}+\beta n^p-\beta (n-1)^p+R(n)-R(n-1)\\
=\alpha (p+1)n^p-\alpha \frac{(p+1)p}2n^{p-1}+\beta pn^{p-1}+Q(n)$$ where $Q$ is of degree at most $p-2$.
By identification, 
$$\alpha =\frac1{p+1},\\\beta =\frac12$$ (which are the first two Faulhaber coefficients.)
This justifies the $\dfrac12$ in a).
For b), consider 
$$S(2n)-2^pS(n)=\frac{2^{p+1}n^{p+1}-2^pn^{p+1}}{p+1}+Q'(n)=\frac{2^pn^{p+1}}{p+1}+Q'(n)$$ 
to get the sum of the powers of the odd integers.