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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 $\lim\limits_{n\to\infty}\frac{\sum_{k=1}^n k^m}{{m+1}}$.

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Please use LaTex for your question. – Thomas Andrews May 27 '12 at 14:26
See this related question. – TMM May 27 '12 at 14:27
Thx Thomas Andrews to LaTex it for me! – ᴊ ᴀ s ᴏ ɴ May 27 '12 at 14:32
I've edited the question to LaTeX, but there was a clear error in your original question which I left in - you ask if: $$\lim_{n\to\infty}\frac{\sum_{k=1}^n k^m}{n^{m+1}} = \frac 1 {n^{m+1}}$$ But the left hand side is a limit as $n$ varies, and the right hand side still has $n$. I think the right hand side should be $\frac 1{m+1}$ – Thomas Andrews May 27 '12 at 14:33
up vote 2 down vote accepted

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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This is a valid answer, and will probably be of use to more advanced readers. However, the question was tagged algebra-precalculus. Another answer utlizing calculus would be to approximate the Riemann integral by$$\int_0^1x^m\,\mathrm{d}x =\lim_{n\to\infty}\sum_{k=1}^n \left(\frac{k}{n}\right)^m\frac1n$$ – robjohn Aug 11 '12 at 1:20
Can you please further explain about $\sum_{k=1}^{n}k^m\leq \int_{1}^{n+1}k^mdk$ ? – ᴊ ᴀ s ᴏ ɴ Aug 11 '12 at 4:20
Is it clear that $k^m \leq \int_k^{k+1} x^m dx$? This is because when $k\leq x \leq k+1$, $k^m \leq x^m$. Integrate this to get $k^m \leq \int_k^{k+1} x^m dx$. Sum this over $k$ from $1$ to $n$ to get the inequality. Then replace $x$ with $k$. – nayrb Aug 11 '12 at 12:17
Very simple, thanks. – ᴊ ᴀ s ᴏ ɴ Aug 12 '12 at 14:46
Evidently, I was wrong about the usage of calculus in a problem tagged algebra-precalculus. – robjohn Aug 12 '12 at 15:55

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 $m<M$, 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 $m<M$, each $\displaystyle\left(\sum_{k=1}^nk^m\right)$ on the right-hand side of $(4)$ is a polynomial in $n$ of degree $m+1<M+1$. 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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Does $O$ notation apply to precalc? – Pedro Tamaroff May 27 '12 at 17:35
Thank you! I get it now! – ᴊ ᴀ s ᴏ ɴ May 28 '12 at 9:08
@Peter: I don't know in what order things are taught in various places, but limits and big-O notation are very closely related. – robjohn May 28 '12 at 13:41

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.

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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 $${1\over n^{m+1}}\sum_{k=1}^n k^m = {1\over m+1} + {1\over n^{m+1}}P_m(n) = {1\over m+1} + O(1/n). $$

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