<s>[INST] Find limits of sequences 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).
 [/INST] Solution for (b):
$$\lim_n\frac{1^p+3^p+\cdots +(2n-1)^p}{n^{p+1}}=\lim_n \frac{1^p+2^p+\cdots +(2n)^p}{n^{p+1}}-\lim_n\frac{2^p+4^p+\cdots+(2n)^p}{n^{p+1}}$$
$$=2^{p+1}\lim_n\frac{1^p+2^p+\cdots +n^p}{n^{p+1}}-2^p\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}$$
$$=(2^{p+1}-2^p).\lim_n\frac{1^p+2^p+\cdots+n^p}{n^{p+1}}=2^p.\lim_n\frac{1}{n}\sum_{r=1}^n \left(\frac{r}{n}\right)^p$$
$$=2^p.\int_0^1x^p\,dx=\frac{2^p}{p+1}$$
</s>