# 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 $$\lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{a_n-a_{n-2}}{b_n-b_{n-2}}$$?

• Why don't you just use $a_n = 1^k + 3^k + \cdots + (2n+1)^k$?
– user99914
Nov 19 '15 at 23:00
• $a_n-a_{n-1} = -0^k + 1^k -2^k + 3^k - 4^k + \cdots -(2n)^k + (2n+1)^k$ does it really help? For me it looks like bigger mess
– Andy
Nov 19 '15 at 23:05
• Note that you do not have even term in every $a_n$.... $a_1 = 1$, $a_2 = 1 + 3^k$, $a_3 = 1+ 3^k + 5^k$......
– user99914
Nov 19 '15 at 23:06
• Damn! I am so stupid, of course $a_n-a_{n-1} \ne -0^k + 1^k -2^k + 3^k - 4^k + \cdots -(2n)^k + (2n+1)^k$ Thanks!
– Andy
Nov 19 '15 at 23:13

Stolz-Cesaro will work nicely here. We want the limit of

$$\frac{(2n+3)^k}{(n+1)^{k+1}-n^k}.$$

All we need is an estimate on the denominator. But by the MVT this equals

$$\tag 1 (k+1)(c_n)^k\cdot 1,$$

for some $c_n \in (n,n+1).$ Now $(1)$ lies between $(k+1)n^k$ and $(k+1)(n+1)^k.$ The rest is easy with the Squeeze theorem.

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,

$\lim_{n \to \infty}{\frac{1^k+3^k+5^k+\cdots+(2n+1)^k}{n^{k+1}}} = \lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+(2n+1)^k}{n^{k+1}}} - 2^k \lim_{n \to \infty}{\frac{1^k+2^k+3^k+\cdots+n^k}{n^{k+1}}}$

$= \lim_{n \to \infty}\frac{(2n+1)^{k+1}}{(k+1)n^{k+1}}-2^k\frac{n^{k+1}}{(k+1)n^{k+1}}$.

$= \lim_{n \to \infty}\frac{2^{k+1} (n+1/2)^{k+1}-2^kn^{k+1}}{(k+1)n^{k+1}}$.

$= \lim_{n \to \infty}\frac{2^{k+1} n^{k+1}-2^kn^{k+1}}{(k+1)n^{k+1}}$.

$= \lim_{n \to \infty}\frac{2^{k+1}-2^k}{k+1}$.

$=\frac{2^k}{k+1}$.