2

I'm trying to get the limit of $\lim\limits_{n \to \infty} \frac{\sum\limits_{k=1}^{n} k^m}{n^{m+1}}$

I know that $\sum\limits_{k=1}^{n} k^m$ is a polynom of $\deg(m +1)$ by the Euler-MacLaurin formula. And that I may apply L'hôspitals rule $m+1$ times.

So there will be some constant in the denominator and the numerator I guess? Maybe something like $(m+1)!$

But I can't find out what exactly yields the $m+1$ the derivitive of that sum. Has anybody a tipp on how to solve this?

Edit: I forgot the $+1$, sry

Algebruh
  • 600

4 Answers4

4

Hint:

$$\frac{1}{n}\sum_{k=1}^n\left( \frac{k}{n} \right)^m \approx \int_{0}^{1}x^m\text{d}x $$

The above is true due to Riemann sum & how we define the integral. Therefore: $$ \lim_{n \to \infty } \frac{1}{n} \sum_{k=1}^{n} \left( \frac{k}{n} \right)^m= \int_{0}^{1}x^m\text{d}x= \frac{1}{m+1} $$

Marek Kryspin
  • 1,052
  • It is not just an approximation. The limit is the Riemann integral –  Sep 03 '20 at 13:27
  • That's what I mean by $\approx$. When $n\to\infty$ then there will be equality. – Marek Kryspin Sep 03 '20 at 13:31
  • So the idea is to reorganize that equation until I have something like $\sum\limits_{k=1}^{n} k^m = \frac{1}{m +1}m^mn$? (Hope I didnt fail somewhere on the way) – Algebruh Sep 03 '20 at 14:07
  • 1
    No, that's not true. Note that you are counting the left side with $n$ going to infinity so: $$\frac{1}{n}\sum_{k=1}^n\left( \frac{k}{n} \right)^m \to \int_{0}^{1}x^m\text{d}x$$ when $n\to \infty$. Therefore you just need to calculate the integral on the right – Marek Kryspin Sep 03 '20 at 14:18
  • Ahh...now I got it...thanks :) – Algebruh Sep 03 '20 at 14:43
1

Applying Stolz–Cesàro theorem, where $a_n=\sum\limits_{k=1}^n k^m$ and $b_n=n^{m+1}$ gives $$\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\\ \frac{(n+1)^m}{(n+1)^{m+1}-n^{m+1}}=\\ \frac{(n+1)^m}{(n+1-n)\left((n+1)^{m}+(n+1)^{m-1}n+(n+1)^{m-2}n^2+...+(n+1)n^{m-1}+n^m\right)}=\\ \frac{(n+1)^m}{(n+1)^{m}+(n+1)^{m-1}n+(n+1)^{m-2}n^2+...+(n+1)^2n^{m-2}+(n+1)n^{m-1}+n^m}=\\ \frac{1}{1+\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+...+\frac{n^{m-2}}{(n+1)^{m-2}}+\frac{n^{m-1}}{(n+1)^{m-1}}+\frac{n^{m}}{(n+1)^{m}}}\to\frac{1}{m+1}, n\to\infty$$ Thus $$\frac{a_n}{b_n}\to\frac{1}{m+1}, n\to\infty$$

Mittens
  • 39,145
rtybase
  • 16,907
0

Let $s(m,n) = \sum_{k=1}^n k^m$, then as you write, $s(n,m)$ is a polynomial in $n$ of degree $m+1$. Thus, $$ \frac{s(m,n)}{n^m} = \Theta(n) \to \infty $$ as $n \to \infty$.

A much more interesting question, of course, would we to explore the asymptotics of $$ \frac{s(m,n)}{n^{m+1}} = \Theta(1). $$ I would start at writing out $s(n,1), s(n,2), \ldots$ and contemplating the sequence of leading coefficients.

gt6989b
  • 54,422
0

You can get more than the limit using generalized harmonic numbers $$ \sum\limits_{k=1}^{n} k^m=H_n^{(-m)}$$ Using asymptotics $$n^{-m} \sum\limits_{k=1}^{n} k^m=n^{-m}\,H_n^{(-m)}=n^{-m} \left(n^m \left(\frac{n}{m+1}+\frac{1}{2}+\frac{m}{12 n}+O\left(\left(\frac{1}{n}\right)^3\right)\right)+\zeta (-m)\right)$$ Expand and get $$\frac{n}{m+1}+\frac{m}{12 n}+\frac{1}{2}\to \infty$$

Claude Leibovici
  • 260,315