Prove $\sum_{i=1}^{n}i^r = \frac{n^{r+1}}{r+1}+\frac{n^r}{2}+P_r(n)$ Such that $P_r$ is a polynomial with degree less than $r$

Question

Question :

For each arbitrary constant $n$, using induction, Prove that :

For each natural number $r$ :

$\sum_{i=1}^{n}i^r = \frac{n^{r+1}}{r+1}+\frac{n^r}{2}+P_r(n)$

Such that $P_r$ is a polynomial with degree less than $r$.

Note : It has to be related with integration but i don't know how show the connection between that polynomial and the rest of it... Also, Induction on which variable?

Thanks in advance.

That is a straightforward consequence of Faulhaber's formula (https://en.wikipedia.org/wiki/Faulhaber's_formula). — Jack D'Aurizio, Dec 28 '16 at 17:31

score 3 · Accepted Answer · answered Dec 28 '16 at 16:00

The problem is an algebraic one and does not really require the machinery of calculus. Below is an easy proof based on induction.

We can prove using induction on $r$ that for all natural numbers $r$ we have $$\sum_{k = 1}^{n}k^{r} = S_{r}(n)$$ where $S_{r}(n)$ is a polynomial of degree $r + 1$ in $n$. Clearly for $r = 0$ we have $S_{0}(n) = n$ so the induction hypothesis is true for $r = 0$. Assume it is true for all $r = 0, 1, 2, \ldots, m - 1$ and then see what happens when $r = m$.

We have via Binomial Theorem $$k^{m + 1} - (k - 1)^{m + 1} = \binom{m + 1}{1}k^{m} - \binom{m + 1}{2}k^{m - 1} + \cdots$$ And summing for $k = 1, 2, \ldots, n$ we get $$n^{m + 1} = (m + 1)S_{m}(n) - \binom{m + 1}{2}S_{m - 1}(n) + \cdots$$ We have now assumed that $S_{r}$ is a polynomial of degree $r + 1$ for all $r = 0, 1, 2, \ldots, m - 1$ and hence from the last equation we see that $$S_{m}(n) = \frac{n^{m + 1}}{m + 1} + \frac{m}{2}S_{m - 1}(n) - \cdots$$ and it follows that $S_{m}(n)$ is of degree $m + 1$ in $n$. It follows via induction that $S_{r}(n)$ is a polynomial of degree $r + 1$ in $n$. Also note that the same proof above shows that the leading term in $S_{r}(n)$ is $n^{r + 1}/(r + 1)$. And because of this we can see that $$S_{r + 1}(n) = \frac{n^{r + 1}}{r + 1} + \frac{r}{2}S_{r - 1}(n) - \cdots = \frac{n^{r + 1}}{r + 1} + \frac{r}{2}\left(\frac{n^{r}}{r} + \frac{r - 1}{2}S_{r - 2}(n) -\cdots\right) - \cdots$$ and finally we get $$S_{r}(n) = \frac{n^{r + 1}}{r + 1} + \frac{n^{r}}{2} + P_{r}(n)$$ where $P_{r}(n)$ is a polynomial of degree less than $r$.

score 2 · Answer 2 · edited Apr 13 '17 at 12:20

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As I showed in this answer, let $f(n,r)$ be equal to your sum whenever $n,r\in\mathbb N$ and let $f(n,0)=n$. Then, we have the following recursive formula:

$$f(x,p)=a_px+\int_0^xf(t,p-1)dt$$

where

$$a_p=1-p\int_0^1f(t,p-1)dt$$

Thus, if $f(x,1)$ is a polynomial of degree $2$ of that form, then it follows by induction that $\sum_{i=1}^ni^r$ is a polynomial of that form.

edited Apr 13 '17 at 12:20

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answered Dec 28 '16 at 15:34

Simply Beautiful Art

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Your technique is nice and very elegant.+1 – Paramanand Singh Dec 28 '16 at 16:04
@ParamanandSingh Thanks. Very much proud that I managed to think it up, and now its being useful :D – Simply Beautiful Art Dec 28 '16 at 16:05
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The joy of mathematical discovery is a rare gift even if the thing discovered is not earth shattering. Enjoy it as much as you can. :) – Paramanand Singh Dec 28 '16 at 16:07
@ParamanandSingh Thank you so much, and I will. – Simply Beautiful Art Dec 28 '16 at 16:08

score 1 · Answer 3 · answered Dec 28 '16 at 17:39

If $r\in\{0,1,2\}$ the claim is trivial, hence we may assume $r>2$ WLOG.
By induction, it is enough to bound the degree of: $$ q_r(n)=\frac{(n+1)^r+n^r}{2}-\frac{(n+1)^{r+1}-n^{r+1}}{r+1}.$$ We have that: $$ [x^{r+1}]q_r(x)=0,\qquad [x^r]q_r(x) = 1-1 = 0$$ $$[x^{r-1}]q_r(x) = \frac{r}{2}-\frac{1}{r+1}\binom{r+1}{2} = 0$$ hence the degree of $q_r(n)$ is at most $r-2$ and the claim easily follows.
It is also a straightforward consequence of Faulhaber's formula.

Prove $\sum_{i=1}^{n}i^r = \frac{n^{r+1}}{r+1}+\frac{n^r}{2}+P_r(n)$ Such that $P_r$ is a polynomial with degree less than $r$

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