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Start with the identity

$\sum_{i=1}^n i^3 = \left( \sum_{i = 1}^n i \right)^2 = \left(\frac{n(n+1)}{2}\right)^2$.

Differentiate the left-most term with respect to $i$ to get

$\frac{d}{di} \sum_{i=1}^n i^3 = 3 \sum_{i = 1}^n i^2$.

Differentiate the right-most term with respect to $n$ to get

$\frac{d}{dn} \left(\frac{n(n+1)}{2}\right)^2 = \frac{1}{2}n(n+1)(2n+1)$.

Equate the derivatives, obtaining

$\sum_{i=1}^n i^2 = \frac{1}{6}n(n+1)(2n+1)$,

which is known to be correct.

Is there any neat reason why this method happens to get lucky and work for this case?

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Well, the leading coefficient on both sides is guaranteed to agree. Past that, the corresponding statement is false if 3 is replaced by 1 and probably also if it's replaced by 2. – Qiaochu Yuan Nov 3 '10 at 16:25
Looks like a valid Lucky Larry entry... :) – J. M. Nov 3 '10 at 16:26
Correction: do it with sum of squares and you get $\sum i = \frac{1}{12}(6n^2+6n+1)$. – Arturo Magidin Nov 3 '10 at 16:27
@Arturo: That's how one of my friends proved $\frac{\sin{x}}{n}=6$ by cancelling $n$ – anonymous Nov 3 '10 at 16:48
So I don't actually know or understand why this works, but I remember this trick working if you start out with the sum of the odd powers, to get the formula the sum of the even powers. When going from the sum of the even powers to the odd powers, you need to add a correction term, which always happens to be a constant (this sequence turns out to be the Bernoulli numbers). – Braindead Nov 3 '10 at 16:53
up vote 24 down vote accepted

Let $f_k(n)=\sum_{i=1}^n i^k$. We all know that $f_k$ is actually a polynomial of degree $k+1$. Also $f_k$ can be characterised by the two conditions: $$f_k(x)-f_k(x-1)=x^k$$ and $$f_k(0)=0.$$ Differentiating the first condition gives $$f_k'(x)-f_k'(x-1)=k x^{k-1}.$$ Therefore the polynomial $(1/k)f_k'$ satisfies the first of the two conditions that $f_{k-1}$ does. But it may not satisfy the second. But then $(1/k)(f_k'(x)-f_k'(0))$ does. So $$f_{k-1}(x)=\frac{f_k'(x)-f_k'(0)}k.$$

The mysterious numbers $f_k'(0)$ are related to the Bernoulli numbers, and when $k\ge3$ is odd they obligingly vanish...

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Oho! That's a very nice proof. – Willie Wong Nov 3 '10 at 17:42
Fantastic. I'm very pleased to see there's something interesting going on in the background here. The same goes for your answer, Willie Wong. – Zach Conn Nov 3 '10 at 19:24

Consider Bernoulli's formula which gives

$$ \sum_1^n i^m = \frac{1}{m+1} \sum_0^m {m+1 \choose k} B_k n^{m+1-k} $$

where $B_k$ is the k'th Bernoulli number. Take the derivative with respect to $n$ on the right hand side, you get

$$ \frac{d}{dn} \sum_1^n i^m = \frac{1}{m+1} \sum_0^m \frac{(m+1)\cdot m!}{(m+1 - k)\cdot (m -1 + 1 - k)! k!}\cdot (m+1-k) B_k n^{m-1 + 1-k} $$

$$= \sum_0^m { m \choose k} B_k n^{m-k} = m \sum_1^n i^{m-1} + B_m$$

So the difference between formally taking the derivative w.r.t. $i$ as you did, and formally taking the derivative w.r.t. $n$, is exactly the Bernoulli constant for the power $m$. Now Bernoulli's constant is equal to 0 for all odd $m$ that is greater than 1. For those cases your observation is also true: the two "derivatives" are equal.

Now, some random guessing, since I am not an analytic number theorist. The deep connection to Bernoulli numbers probabaly has something to do with analytic number theory and harmonic analysis that I do not know myself. That the formulae work for odd powers is possibly something related to the method of descent, where analytical results in even dimensions can be arrived from results in one higher dimension via some sort of trace.

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This is quite obvious if expressed most naturally in terms of the Bernoulli polynomials, viz.

$\quad\quad\quad\quad\rm\ \ \ \sum n^k\ = \:\frac{1}{k+1} (B_{k+1}(n+1) - B_{k+1}(0))\ $

and $\rm\quad\quad\ B_k^{\:'}(x)\ =\ k\ B_{k-1}(x)\ $

thus $\ \rm\ (\sum n^k)'\ =\ k\ \sum n^{k-1} + B_{\:k} $

According to Knuth's exposition these identities go all the way back to Jacobi - with special cases known to Faulhaber. Note that the identities in Robin Chapman's post are simply special cases of these well-known identities for Bernoulli polynomials.

The Bernoulli polynomials are a special-case of polynomials amenable to study by the powerful techniques of the Umbral Calculus. This is the best way to understand the genesis of their many interesting properties. For a nice introduction see Steven Roman's book.

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