Limit of $\lim\limits_{n\to\infty}\frac{\sum_{m=0}^n (2m+1)^k}{n^{k+1}}$ I wanted to find the limit of: ($k \in N)$
$$\lim_{n \to \infty}{\frac{1^k+3^k+5^k+\cdots+(2n+1)^k}{n^{k+1}}}.$$
Stolz–Cesàro theorem could help but $\frac{a_n-a_{n-1}}{b_n-b_{n-1}}$ makes big mess here:
$$\lim_{n \to \infty}{\frac{-0^k+1^k-2^k+3^k-4^k+5^k-6^k+\cdots-(2n)^k+(2n+1)^k}{n^{k+1}-(n-1)^{k+1}}}.$$
Is following statement true as well $$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_n-a_{n-2}}{b_n-b_{n-2}}$$?

Using Faulhaber's formula, 
$\lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+n^k}{n^{k+1}}} = \frac{1}{k+1}$.
Then,
$\lim_{n \to \infty}{\frac{1^k+3^k+5^k+\cdots+(2n+1)^k}{n^{k+1}}} = \lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+(2n+1)^k}{n^{k+1}}} - 2^k \lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+n^k}{n^{k+1}}}$
$ = \lim_{n \to \infty}\frac{(2n+1)^{k+1}}{(k+1)n^{k+1}}-2^k\frac{n^{k+1}}{(k+1)n^{k+1}}$. 
$ = \lim_{n \to \infty}\frac{2^{k+1} (n+1/2)^{k+1}-2^kn^{k+1}}{(k+1)n^{k+1}}$. 
$ = \lim_{n \to \infty}\frac{2^{k+1} n^{k+1}-2^kn^{k+1}}{(k+1)n^{k+1}}$. 
$ = \lim_{n \to \infty}\frac{2^{k+1}-2^k}{k+1}$. 
$ =\frac{2^k}{k+1}$. 
