# Proving the odd Bernoulli numbers are zero

suppose we define the Bernoulli numbers $b_n, n = 1, 2, 3, \ldots$ by the Faulhaber's fomula

$$\begin{eqnarray*}1 + 2^k + 3^k + \ldots n^k &=& \frac{1}{k+1}[n^{k+1} + b_1 c(k+1,2) n^k + b_2 c(k+1,3)n^{k-1} + \ldots ]\\&=& \frac{(n+b)^{k+1}-b^{k+1}}{k+1}\end{eqnarray*}$$

with the proviso we replace $b^k$ by $b_k$ in the binomial expansion and $c(k,l)$ is $k$ choose $l.$

my question is, how do you show all the odd Bernoulli numbers except $b_1$ is zero without invoking heavy analytical tools so that the reason can be explained to a student who has only had one or two semesters of calculus in high school.

-
this is exactly what you don't want - but I think that first verifying that the function $f(x)=\sum x^k/k!$ satisfies $f(x)f(y)=f(x+y)$ and then doing the usual stuff ($\sum_k (x^k+(2x)^k+\dots+(nx)^k)/k!$) is clean and has an added value –  user8268 May 5 '11 at 14:23
I'm not sure what heavy analytical tools means. My argument below uses Taylor series (second term of Calculus, as I recall high school) and (implicitly) that you can manipulate Taylor series like polynomials. –  David Speyer May 5 '11 at 14:26
by heavy machinery i included the generating functions $\frac{t}{1 -e^{-t}}$ and $\frac{x}{2}{\coth(\frac{x}{2}}).$ we will get both of them as simple consequence of euler-maclaurin formulae. faulhaber's is a preamble for the euler formule. the students need some thing more interesting math after the ap exams, i am trying to show them how far you can go with integration by parts. –  abel May 5 '11 at 14:45
this is not integration by parts. –  quanta May 5 '11 at 15:25

Define $\sigma_k(n) = \sum_{i=1}^n i^k$ then since $\sigma_1(n) = \frac{n(n+1)}{2}$

First to show that the coefficient of $n$ in $\sigma_{2k+1}$ is $b^{2k+1}$: It's immediate from applying the binomial theorem to $\sigma_{2k+1}(n) = \frac{(n+b)^{k+1}-b^{k+1}}{k+1}$. Then we can show this coefficient is always zero by showing these polynomials are divisible by something with $n^2$ as its lowest power.

This argument is due to Pascal (I could not possibly come up with it!) taken from here,

$$\begin{array} . \sigma_1(n)^k &=& \sum_{m=1}^n \left[ \left(\frac{m(m+1)}{2}\right)^k - \left(\frac{(m-1)m}{2}\right)^k \right] \\ &=& \sum_{m=1}^n \sum_{r=0}^k \binom{k}{r} \left(\frac{m}{2}\right)^k m^r (1 - (-1)^{k-r}) \\ &=& \frac{1}{2^k} \sum_{r=0}^k \binom{k}{r} \sigma_{k+r}(n) (1 - (-1)^{k-r}) \end{array}$$

Now consider $k$ odd, many of the terms cancel, we are left with the fact that $\sigma_1^k$ is a linear combination of various $\sigma_{i}$ for $i$ odd.

The relation $\sigma_3 = \sigma_1^2$ is classical and can be proved in a variety of simple ways. Induction on (odd) $k$ shows that every other odd one ($\sigma_5$, $\sigma_7$ and so on) is also divisible by $\sigma_1^2$. In fact you can try out these induction arguments concretely for given $k$ to express these summations in terms of triangular numbers. That is called producing the Faulhauber polynomials.

This proves that there is no $n$ terms in these odd ($\ge 3$) sums of powers, but since the coefficient of $n$ in $\sigma_{2k+1}$ is $b_{2k+1}$ this proves that they are all zero.

-
quanta, i will try to understand your proof. is it like expanding $(1 + 2 + \ldots +n)^k?$ –  abel May 5 '11 at 15:33
@abel, it's not my proof but I understood it so feel free to ask any questions about bits that don't make sense. I don't know what you are referring to with that question though. –  quanta May 5 '11 at 15:54
Why sigma_1^k = that sum? Well the terms of the summation are like (a - b) + (b - c) + (c - d) + ... + (x - y) and y = 0 so it's just a = n(n+1)/k. (This is called telescoping because it's like folding up a telescope) –  quanta May 5 '11 at 15:56

Why not take the opportunity to expose them to the glory of generating functions?

Define the Bernouli numbers to be the coefficients in $$\frac{t}{1-e^{-t}} = \sum B_m \frac{t^m}{m!}.$$

Prove that these give the sum of the $k$-th powers by equating coefficients of $x^k$ in $$\sum_{j=1}^n \frac{j^k x^k}{k!} = \sum_{j=1}^n e^{jx} = \frac{e^{(n+1)x} - e^x}{e^x-1}$$ $$= \left( \frac{e^{nx} -1}{x} \right) \left( \frac{x}{1-e^{-x}} \right) = \sum_{\ell=0}^{\infty} \frac{n^{\ell+1} x^{\ell}}{(\ell+1)!} \cdot \sum B_m \frac{t^m}{m!}.$$

Then prove that the odd Bernoulli numbers vanish by noting that $$\sum B_m \frac{t^m}{m!} - \sum B_m \frac{(-t)^m}{m!} = \frac{t}{1-e^{-t}} - \frac{-t}{1-e^t} = \frac{t (e^t-1)}{e^t-1} = t.$$

-
david, i did not want to define bernoulli number with the generating function you give above for the reason that it will mysterious how anyone can come up with such. for ease of introduction i chose the jacob bernoullis. –  abel May 5 '11 at 16:34
What's with the change of name? –  Pedro Tamaroff Jul 27 '13 at 23:32
"...please downvote any answers which I leave between 9 AM and 5 PM on weekdays, Eastern Time." This is interesting...! –  Pedro Tamaroff Jul 27 '13 at 23:33

thanks for all the responses. i will collect the results. this is too large to fit into a comment to quanta's response.

we will use $\sigma_k = 1^k + 2^k + \ldots n^k$ proving the odd bernoulli constants $b_k$ are zero for $= k = 3, 5, \ldots.$ let us fix the notation and collect some elementary results. $\sigma_k = 1^k + 2^k + \ldots n^k, \sigma_1 = \frac{n(n+1)}{2}, \sigma_3 = \sigma_1^2.$

the idea of the proof is
to show that $n^2$ divides $\sigma_k$ for all $k \ge 3$ by induction. clearly true for $k = 3.$ we will show that $\sigma_k$ is a linear combination of $\sigma_{k-2}, \ldots, \sigma_3$ and $\sigma_1^k$ for all $k \ge 3.$ that concludes the proof.

now i will establish the needed linear dependence of $\sigma_k$ which is $$\sigma_1^k = \frac{1}{2^{k-1}} \cases{[ {k \choose 1}\sigma_{2k-1} + {k \choose 3}\sigma_{2k-3} +\ldots +\sigma_k ] & for k odd, \cr [ {k \choose 1}\sigma_{2k-1} + \ldots + {k \choose 3}\sigma_{2k-3} ] & for k even.}$$ i will show all the intermediate steps for $k = 5.$ it is easy to see the steps for the $k = 5$ carry over to general case. \ $\begin{array} \sigma_1^5 = (1+2\ldots n)^5 = \frac{n^5(n+1)^5}{2^5} \\ ~~~~ = \frac{1}{2^5}\{[n^5(n+1)^5 -(n-1)^5n^5] + [(n-1)^5n^5 -(n-2)^5(n-1)^5] +\ldots +[2^53^5 - 1^52^5] + 1^5 2^5 \}\\ ~~~~ = \frac{1}{2^5}\{ n^5[(n+1)^5 -(n-1)^5] + ((n-1)^5[n^5 -(n-2)^5] +\ldots +2^5[3^5 - 1^5] + 1^5 2^5 \}\\ ~~~~ = \frac{1}{2^5}\{ 2n^5[{5 \choose 1}n^4 + {5 \choose 3}n^2 + 1] + 2(n-1)^5[{5 \choose 1}(n-1)^4 + {5 \choose 3}(n-3)^2 + 1]+\\ ~~~~~~ \ldots + 2^5[{5 \choose 1}2^4 + {5 \choose 3}2^2 + 1] + 1^5[{5 \choose 1}1^4 + {5 \choose 3}1^2 + 1] \}\\ ~~~~~~ = \frac{1}{2^4}\{ {5 \choose 1}[n^9 + (n-1)^9 + \ldots + 2^9 + 1^9] + {5 \choose 3}[n^7 + (n-1)^7 + \ldots + 2^7 + 1^7] + [n^5 + (n-1)^5 + \ldots + 2^5 + 1^5] \}\\ ~~~~~ = \frac{1}{2^4}[ {5 \choose 1}\sigma_9 + {5 \choose 3}\sigma_7 + \sigma_5 ]. \end{array}$

-