Computing the Limit of a Power Series

[Julio1]

Compute $\displaystyle\lim_{n\to +\infty}\dfrac{1^p+2^p+3^p+\cdots +n^p}{n^{p+1}}.$

Spoiler

Hello!, how it can calculate this limit? Thanks :)

 

[mathbalarka]

I am not sure what techniques you are familiar with for calculating limits, but this seems to be a job for Stolz-Cesaro theorem :

$$\begin{aligned}\lim_{n \to \infty} \frac{1^p + 2^p + \cdots + n^p}{n^{p+1}} &= \lim_{n \to \infty} \frac{\left ( 1^p + 2^p + \cdots + (n+1)^p \right ) - \left( 1^p + 2^p + \cdots + n^p \right)}{(n+1)^{p+1} - n^{p+1}} \\ &= \lim_{n \to \infty} \frac{(n+1)^p}{(n+1)^{p+1} - n^{p+1}} \\ & = \lim_{n \to \infty} \frac{(n+1)^p}{(p+1) \cdot n^p + \mathcal{O}(n^{p-1})} \\ &= \lim_{n \to \infty} \frac{1}{p+1} \cdot \frac{1}{1+o(1)} = \boxed{\dfrac1{p+1}}\end{aligned}$$

Where $\mathcal{O}$ and $o$ are asymptotic notations. I am guessing that there might be some elementary way to do to this, so am interested in other solutions.

 

[Euge]

Compute $\displaystyle\lim_{n\to +\infty}\dfrac{1^p+2^p+3^p+\cdots +n^p}{n^{p+1}}.$

Spoiler

Hello!, how it can calculate this limit? Thanks :)

I'm assuming $p$ is a real number different from $-1$. The limit, which can be written

\(\displaystyle \lim_{n\to +\infty} \frac{1}{n}\sum_{k = 1}^n \left(\frac{k}{n}\right)^p,\)

is the limit of a sequence of Riemann sums for the function $f(x) = x^p$ over the interval $[0,1]$. So it has value

\(\displaystyle \int_0^1 x^p \, dx.\)

Compute the integral to get the result.