Finding the general/closed form of $\sum_{k=1}^n k^a$

Question

I recently noticed the following:

$$\sum_{i=1}^ni=\frac{n(n+1)}{2}$$ $$\sum_{i=1}^{n}i^2 = \dfrac{n(n+1)(2n+1)}{6} = \Bigg(\sum_{i=1}^ni\Bigg) \cdot \frac{2n+1}{3}$$ $$\sum\limits_{i = 1}^n i^3 = \Bigg( \frac{n(n+1)}{2}\Bigg)^2 = \Bigg(\sum_{i=1}^ni\Bigg)^2$$ $$\sum_{i=1}^ni^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30} = \Bigg(\sum_{i=1}^ni^2\Bigg) \cdot \frac{3 n^2 + 3 n - 1}{5}$$

It seems that the sums build on each other in some interesting way, and I'm interested in finding a general form. A cursory investigation found that $$\sum_{k=0}^n k^a = H_n^{(-a)}$$ where $H_n^{(-a)}$ is the $n$th hyperharmonic number of $r$th order. However, I doubt this would help because this (a) is sort of circular given the definition of $H_n$, (b) has no closed form, and (c) doesn't reveal the multiplicative connection among the terms.

Is it possible to derive a closed form for $\sum_{k=0}^n k^a$? Why or why not? And, of course, if one exists, what is it?

Answer 1

The Summa Potestatum involves many other relations.
Some of the more interesting are

Recursion

$$ \bbox[lightyellow] { S_m (n) = \sum\limits_{0\, \le \,k\, \le \,n - 1} {k^{\,m} } = \left[ {1 \le n} \right]\left( {\left[ {0 = m} \right] + \sum\limits_k {\left( \matrix{ m \cr k \cr} \right)S_k (n - 1)} } \right)\quad \left| {\;0 \le {\rm integer }m,n} \right. } \tag{1}$$ where $[P]$ denotes the Iverson bracket

Relation with Eulerian, Stirling and Bernoulli Numbers

$$ \bbox[lightyellow] { \eqalign{ & S_m (n) = \sum\limits_{0\, \le \,k\, \le \,n - 1} {k^{\,m} } \quad \left| {\;0 \le {\rm integer }m,n} \right. = \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {\left\langle \matrix{ m \hfill \cr j \hfill \cr} \right\rangle \left( \matrix{ n + j \cr m + 1 \cr} \right)} = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {\;j!\;\left\{ \matrix{ m \cr j \cr} \right\}\left( \matrix{ n \cr j + 1 \cr} \right)} = \cr & = {1 \over {m + 1}}\sum\limits_{0\, \le \,j\, \le \,m} {\left( \matrix{ m + 1 \cr j \cr} \right)\;B_{\,j} \;n^{\,m + 1 - j} } \cr} }\tag{2}$$ where - the angle braces denote the Eulerian Numbers (1st kind) - the curly braces correspond to the Stirling Numbers of 2nd kind - $B_j$ are the Bernoulli Numbers ($B_1=-1/2$).

The first two, at least, use numbers well defined in its own and so avoid the "circularity" you are lamenting.

We can see how for instance it comes that $$ \eqalign{ & S_3 (n) = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {\;j!\;\left\{ \matrix{ 3 \cr j \cr} \right\}\left( \matrix{ n \cr j + 1 \cr} \right)} = \left( {\left( \matrix{ n \cr 2 \cr} \right) + 6\left( \matrix{ n \cr 3 \cr} \right) + 6\;\left( \matrix{ n \cr 4 \cr} \right)} \right) = \cr & = {{n\left( {n - 1} \right)} \over 2} + n\left( {n - 1} \right)\left( {n - 2} \right) + {{n\left( {n - 1} \right)\left( {n - 2} \right)\left( {n - 3} \right)} \over 4} = \cr & = {{n\left( {n - 1} \right)\left( {n^{\,2} - n} \right)} \over 4} = \left( {{{n\left( {n - 1} \right)} \over 2}} \right)^{\,2} = S_1 (n)^{\,2} \cr} $$ The Faulhaber's Formula that Achille hui already indicated, provides much more insight.

Relation with Bernoulli Polynomial and Hurwitz Zeta

Finally it is interesting to note that the Indefinite Sum is $$ \bbox[lightyellow] { \eqalign{ & \sum\nolimits_{\;x\;} {x^{\,m} } + c\quad \left| {\;0 \le {\rm integer }m} \right. = \cr & = \left( {\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {j!\left\{ \matrix{ m \cr j \cr} \right\}\left( \matrix{ x \cr j + 1 \cr} \right)} } \right) = \cr & = \sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m} \right)} {\left\langle \matrix{ m \cr j \cr} \right\rangle \left( \matrix{ x + j \cr m + 1 \cr} \right)} = {1 \over {m + 1}}\sum\limits_{\left( {0\, \le } \right)\,j\,\left( { \le \,m + 1} \right)} {\left( \matrix{ m + 1 \cr j \cr} \right)\;B_j \;x^{\,m + 1 - j} } = \cr & = {1 \over {m + 1}}B(m + 1,x) = - \,\zeta ( - m,x) \cr} } \tag{3}$$ where $B(n,x)$ are the Bernoully polynomials and $\zeta(n,x)$ the Hurwitz Zeta Function.
And it is also possible to establish a connection with the Generalized Harmonic numbers (as you already found) and many others.

Answer 2

So apparently, the "multiplicative" behavior I was looking for is almost certainly a coincidence. Perhaps what I'm seeing is commonality between terms, since the sum I'm looking for is defined by a summation published by Jacob Bernoulli in 1713:

$$\sum_{k=1}^n k^p = \frac{1}{p+1} \sum_{j=0}^p \binom{p+1}{j} B_j n^{p+1-j}$$

where $B_j$ is the $j$th Bernoulli number of the second kind.

So, for instance, if we had $\sum k^4$, we'd note that $$B_0 = 1, B_1 = \frac 12, B_2 = \frac 16, B_3 = 0, B_4 = -\frac {1}{30}$$ and then compute $$\sum_{k=1}^n k^4 = \frac{1}{5} \sum_{j=0}^p \binom{5}{j} B_j n^{5-j}$$ $$=\frac15(B_0 n^5 + 5B_1 n^4 + 10 B_2 n^3 + 10 B_3 n^2 + 5 B_4 n)$$ $$=\frac15n^5 + \frac12 n^4 + \frac 13 n^3 - \frac {1}{30}n$$

Answer 3

Let $$ S_n(p)=\sum_{k=1}^{n} k^p\qquad n, p\in\mathbb N ~~~~~\text{called Cavalieri sum of oder p}$$ Therefore, using the Binomial formula we get $$ (k+1)^p = k^p+ \sum_{i=0}^{p-1}\binom{p}{i} k^i$$ where $\binom{p}{i}= \frac{p!}{i!(p-i)!}$. summing up both side yields, $$\sum_{k=1}^{n} (k+1)^p =\sum_{k=1}^{n} k^p+\sum_{i=0}^{p-1}\binom{p}{i} \sum_{k=1}^{n} k^i = S_n(p) +\sum_{i=0}^{p-1}\binom{p}{i} S_n(i) $$

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

Hence finally we get the formula :

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

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

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