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I have been wanting to understand how to find the sum of this series.

$$1^p + 2^p + 3^p +{\dots} + n^p$$

I am familiar with Gauss' diagonalised adding trick for the sum of the first $n$ natural numbers.

I can prove the formulas for

$$\begin{align} \sum_{1}^{n} k^2 &= \frac{n(n+1)(2n+1)}{6}\\ \sum_{1}^{n} k^3 &= \frac{n^2(n+1)^2}{4}\\ \sum_{1}^{n} k^4 &= \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} \\ \end{align}$$

With mathematical induction. But, beyond that even proofs with mathematical induction are difficult.

I'm interested in learning the theory and the proof behind Faulhaber's formula. What is the knowledge required to understand this proof ?

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The following interesting derivation is from Aigner's "A Course in Enumeration" (Springer, 2007).

Remember that if we have the following exponential generating functions:

$\begin{align} \widehat{A}(z) &= \sum_{n \ge 0} a_n \frac{z^n}{n!} \\ \widehat{B}(z) &= \sum_{n \ge 0} b_n \frac{z^n}{n!} \end{align}$

then also:

$\begin{align} \widehat{A}(z) \cdot \widehat{B}(z) &= \sum_{n \ge 0} \left( \sum_{0 \le k \le n} \frac{a_k}{k!} \frac{b_{n - k}}{(n - k)!} \right) z^n \\ &= \sum_{n \ge 0} \left( \sum_{0 \le k \le n} \frac{n!}{k! (n - k)!} a_k b_{n - k} \right) \frac{z^n}{n!} \\ &= \sum_{n \ge 0} \left( \sum_{0 \le k \le n} \binom{n}{k} a_k b_{n - k} \right) \frac{z^n}{n!} \end{align}$

Let's define:

$\begin{align} S_m(n) = \sum_{1 \le k \le n - 1} k^m \end{align}$

We can define the exponential generating function:

$\begin{align} \widehat{S}_n(z) &= \sum_{m \ge 0} S_m(n) \frac{z^m}{m!} \\ &= \sum_{1 \le k \le n - 1} \sum_{m \ge 0} \frac{k^m z^m}{m!} \\ &= \sum_{1 \le k \le n - 1} \mathrm{e}^{k z} \\ &= \frac{\mathrm{e}^{n z} - 1}{\mathrm{e}^z - 1} \\ \end{align}$

This is almost the exponential generating function of the powers of $n$:

$\begin{align} \widehat{P}_n(z) &= \sum_{m \ge 0} n^m \frac{z^m}{m!} \\ &= \mathrm{e}^{n z} \end{align}$

We can write:

$\begin{align} (\widehat{P}_n(z) - 1) \widehat{B}(z) = z \widehat{S}_n(z) \tag{1} \end{align}$

where we have the exponential generating function of the Bernoulli numbers:

$\begin{align} \widehat{B}(z) &= \frac{z}{\mathrm{e}^z - 1} \\ &= \sum_{n \ge 0} B_n \frac{z^n}{n!} \end{align}$

Comparing coefficients of $z^{m + 1}$ in (1):

$\begin{align} \sum_{m \ge 1} z^m \sum_{0 \le k \le m} \binom{m}{k} \frac{(n z)^{m - k}}{(m - k)!} B_k = \sum_{m \ge 0} S_m(n) \frac{z^{m + 1}}{m!} \end{align}$

we get after simpĺifying:

$\begin{align} S_m(n) = \frac{1}{m + 1} \, \sum_{0 \le k \le m} \binom{m + 1}{k} B_k n^{m + 1 - k} \end{align}$

Note: The formula given is often associated with Faulhaber, but Faulhaber's formulas where quite different (and computationally more efficient). This formula is due to Bernoulli.

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    $\begingroup$ Thank you for the wonderful book recommendation. I am still working through your proof and trying to understand it. $\endgroup$ – user230452 Feb 26 '16 at 4:58
  • $\begingroup$ @user230452, ask if you require any clarification. I'll be happy to update the answer. $\endgroup$ – vonbrand Feb 26 '16 at 11:26
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    $\begingroup$ I'm working through generatingfunctionology to understand what the exponential generating function is and how you applied it here. :) $\endgroup$ – user230452 Feb 26 '16 at 12:46
  • $\begingroup$ Do you know any more good books like the one you mentioned ? $\endgroup$ – user230452 Feb 26 '16 at 13:26
  • $\begingroup$ @user230452, that depends on what you want. And asking for a list of "good books" is off-topic here. But you can search (I try to give references to the places where I found particularly interesting tidbits, for one). $\endgroup$ – vonbrand Feb 26 '16 at 13:34
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For the expression of $\sum n^k$ as a polynomial in $n$ with Bernoulli numbers in the coefficients, the background is calculus of finite differences.

For seeing why and when the linear factors $n$, $n+1$ and $2n+1$ divide the answer, the background needed is some basic algebra of polynomials. Also for showing that the sum for odd $k$ is a polynomial in $n(n+1)$.

The Wikipedia page is informative and links to an expository paper by Beardon on the same subject.

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  • $\begingroup$ What do I need to know and do to understand calculus of finite differences ? What are some good resources ? $\endgroup$ – user230452 Feb 22 '16 at 16:33
  • $\begingroup$ Essentially no background except algebra and familiarity with polynomials. Any book with "calculus of finite differences" as its title should do. It was popular stuff in the 19th century which implies that the necessary prior knowledge is minimal by today's standards. And the old books are quite readable. $\endgroup$ – zyx Feb 22 '16 at 17:43
  • $\begingroup$ @user230452: Or try this: cs.purdue.edu/homes/dgleich/publications/… $\endgroup$ – Giuseppe Negro Feb 22 '16 at 18:50
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This probably won't take you all the way to Faulhaber's formula, but it can help you get to the closed form of $\sum_{k=1}^nk^p$ for any whole $p$.

$$\sum_{k_1=0}^{n-1}\sum_{k_2=0}^{k_0-1}\sum_{k_3=0}^{k_1-1}\dots\sum_{k_x=0}^{k_{x-1}-1}1=\frac{n(n-1)(n-2)\dots(n-x+1)}{1\times2\times3\times\dots\times x}$$

$$=\frac{n!}{x!(n-x)!}=\binom nx$$

For example,

$$\sum_{k=1}^nk^0=1+1+1+\dots+1=\sum_{k=0}^{n-1}1=n$$

$$0^1+\sum_{k=1}^nk^1=\sum_{k=0}^nk^1=\sum_{k_1=0}^{(n+1)-1}\sum_{k_2=0}^{k_1-1}1=\frac{(n+1)((n+1)-1)}2=\frac{n(n+1)}2$$

And you can keep doing this for higher values of $p$.

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You might like to ready my paper https://search.proquest.com/openview/f8786728002514b2de4eaa379d175640/1?pq-origsite=gscholar&cbl=2035960

published in The Mathematical Gazette. It gives a matrix method for producing series of sums of any power. I can send you a copy if you can't find it on line.

Nigel Derby

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  • $\begingroup$ Dear sir, thank you for the reference. You can send it to me at - ghosh_saikat4000@yahoo.co.in $\endgroup$ – user230452 Nov 19 '17 at 15:26
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Here I give an induction formula: from which all sums can be derived:

Let $$ S_n(p)=\sum_{k=1}^{n} k^p\qquad n, p\in\mathbb N ~~~~~\text{called Cavalieri sum of oder p}$$

then, We know the following Binomial formula

$$ (k+1)^p = k^p+ \sum_{i=0}^{p-1}\binom{p}{i} k^i$$ where $\binom{p}{i}= \frac{p!}{i!(p-i)!}$. Which implies that,

$$\sum_{k=1}^{n} (k+1)^p =\sum_{k=1}^{n} k^p+\sum_{i=0}^{p-1}\binom{p}{i} \sum_{k=1}^{n} k^i = S_n(p) +\sum_{i=0}^{p-1}\binom{p}{i} S_n(i) $$

But $$\sum_{k=1}^{n} (k+1)^p = \sum_{k=2}^{n+1} k^p = S_{n+1}(p) -1 = S_n(p) +(n+1)^p -1$$

Hence finally we get the formula :

$$\color{red}{(n+1)^p -1 =\sum_{i=0}^{p-1}\binom{p}{i} S_n(i)} $$

From this it is possible to compute the sum for any $p\ge 1 $ in $ \mathbb N $.

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