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By considering:

$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^1}{n^{2}} = \frac 1 2$ $\lim_{n\to\infty}\frac{\sum_{k=1}^n k^2}{n^{3}} = \frac 1 3$ $\lim_{n\to\infty}\frac{\sum_{k=1}^n k^3}{n^{4}} = \frac 1 4$

Determine if this is true: $\lim_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}} = \frac 1 {{m+1}}$

If it is, prove it.

If it is not, evaluate $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{{m+1}}$.

  • 0
    See http://math.stackexchange.com/questions/478344/what-is-the-result-of-lim-n-to-infty-frac-sumn-i-1-iknk12014-09-28

4 Answers 4

0

What is true is that $\sum_{k=1}^n k^m = {n^{m + 1}\over m+1} + P_m(n),$ where $\deg(P) <= m.$ Dividing you will get ${1\over n^{m+1}}\sum_{k=1}^n k^m = {1\over m+1} + {1\over n^{m+1}}P_m(n) = {1\over m+1} + O(1/n). $

7

Since this is tagged , I will refrain from using the Euler-Maclaurin Sum Formula (which is exact in the case of polynomials).

Let's first show that $ \sum_{k=1}^nk^m\text{ is a polynomial of degree }m+1\text{ in }n\text{ whose lead coefficient is }\frac{1}{m+1}\tag{1} $

Since $\displaystyle\sum_{k=1}^n1=n$, we have shown that $(1)$ is true for $m=0$.

Suppose that $(1)$ is true for all $m, then $ k^{M+1}-(k-1)^{M+1}=\sum_{m=0}^{M}\binom{M+1}{m}(-1)^{M-m}k^m\tag{2} $ If we sum $(2)$ for $k$ from $1$ to $n$, the telescoping sum on the left yields $ n^{M+1}=\sum_{m=0}^{M}\binom{M+1}{m}(-1)^{M-m}\left(\sum_{k=1}^nk^m\right)\tag{3} $ Isolating the $m=M$ term from the right-hand side of $(3)$ gives us $ \sum_{k=1}^nk^M=\frac{1}{M+1}n^{M+1}-\frac{1}{M+1}\sum_{m=0}^{M-1}\binom{M+1}{m}(-1)^{M-m}\left(\sum_{k=1}^nk^m\right)\tag{4} $ Since $(1)$ holds for each $m, each $\displaystyle\left(\sum_{k=1}^nk^m\right)$ on the right-hand side of $(4)$ is a polynomial in $n$ of degree $m+1. Thus, because of the $n^{M+1}$, the right-hand side of $(4)$ is a polynomial in $n$ of degree $M+1$ whose lead coefficient is $\dfrac{1}{M+1}$.

Thus, we have shown $(1)$ for all $m\ge0$.

Consider $ \frac{1}{n^{m+1}}\sum_{k=1}^nk^m\tag{5} $ The sum in $(5)$ is a polynomial in $n$ of degree $m+1$ whose lead coefficient is $\dfrac{1}{m+1}$. Therefore, $(5)$ becomes $ \frac{1}{m+1}+O\left(\frac1n\right)\tag{6} $ Thus, we get that $ \lim_{n\to\infty}\;\frac{1}{n^{m+1}}\sum_{k=1}^nk^m=\frac{1}{m+1}\tag{7} $

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    @Peter: I don't know in what order things are taught in various places, but limits and big-O notation are very closely related.2012-05-28
6

If you are familiar with it, this is a standard application of Stolz-Cesàro.

SC says that this limit is the same as

$\lim_n \frac{(n+1)^m}{(n+1)^{m+1}-n^{m+1}}$

It is an easy exercise (I'll leave this part to you) to prove that this limit is exactly $\frac{1}{m+1}$.

P.S. Stolz-Cesàro is for sequences exactly what l'Hôpital is for functions.

5

This follows easily from a comparison with an integral.

Since $\int_0^{n} k^m dk \leq \sum_{k=1}^n k^m \leq \int_1^{n+1} k^m dk,$ we have $\frac{n^{m+1}}{m+1}\leq \sum_{k=1}^n k^m \leq \frac{(n+1)^{m+1}}{m+1}-\frac{1}{m+1}.$ Divide by $n^{m+1}$ to get $\frac{1}{m+1} \leq \frac{1}{n^{m+1}} \sum_{k=1}^n k^m \leq \frac{1}{m+1} \cdot \frac{(n+1)^{m+1}}{n^{m+1}} - \frac{1}{m+1} \cdot \frac{1}{n^{m+1}}.$ Taking $n \rightarrow \infty$ gives the desired result.

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    Sorry. I'm a poor question- tagger.2012-08-18