I found this nice explicit formula for the Bernoulli numbers:

$$B_n = \sum_{k \mathop = 0}^n \sum_{i \mathop = 0}^k (-1)^i \binom k i \frac {i^n} {k + 1}$$

I can't find a proof though. I want to prove it from the generating function definition:

$$ \frac x {e^x - 1} = \sum_{n \mathop = 0}^\infty B_n \frac {x^n} {n!}$$

Any proof sketches or links will be appreciated.

  • $\begingroup$ That is quite far from a "closed form" in my book... $\endgroup$ – vonbrand May 6 '14 at 12:15
  • $\begingroup$ @vonbrand I agree. I changed it to 'explicit formula'. $\endgroup$ – Superbus May 6 '14 at 12:18
  • $\begingroup$ If you need a reference (no proof): dlmf.nist.gov/24.6.E9 with the subsitution $i\leftrightarrow j$ $\endgroup$ – gammatester May 6 '14 at 13:31
  • $\begingroup$ The expression seems to date back to this article by Worpitzky from 1883. A very good read if you like German papers with ancient notation. $\endgroup$ – Daniel R May 6 '14 at 13:45
  • $\begingroup$ @LuciusTarquiniusSuperbus I have a proof of your question regarding the Bernoulli number asymptotics but the question was deleted before I could answer. $\endgroup$ – Marko Riedel May 6 '14 at 21:46

So as not to get mixed up with complex variables use $j$ rather than $i$ to get $$\sum_{k=0}^n \sum_{j=0}^k (-1)^j {k\choose j} \frac{j^n}{k+1}.$$ This is $$\sum_{k=0}^n \frac{1}{k+1} \sum_{j=0}^k (-1)^j {k\choose j} j^n = \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times {n\brace k}.$$

Recall the classic generating function of the Stirling numbers of the second kind which yields $${n\brace k} = n! [z^n][u^k] \exp(u(\exp(z)-1)).$$ Substituting this into the sum gives $$n![z^n] \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times [u^k] \exp(u(\exp(z)-1)) \\ = n![z^n] \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times k! \times \frac{(\exp(z)-1)^k}{k!} \\= n![z^n] \sum_{k=0}^n \frac{1}{k+1} (-1)^k \times (\exp(z)-1)^k$$ Now observe that $\exp(z)-1$ starts at $z$ and hence we can extend the summation to infinity without affecting $[z^n]$ to get $$n![z^n] \sum_{k=0}^\infty \frac{1}{k+1} (-1)^k \times (\exp(z)-1)^k \\=n! [z^n] \frac{1}{\exp(z)-1} \sum_{k=0}^\infty \frac{1}{k+1} (-1)^k \times (\exp(z)-1)^{k+1} \\= n! [z^n] \frac{1}{\exp(z)-1} \log(1+\exp(z)-1) = n! [z^n] \frac{z}{\exp(z)-1}.$$ Done.

Nice how Bernoulli numbers show up in both analytic number theory and combinatorics.

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  • $\begingroup$ Note: I believe (electrical) engineers make the exact opposite change-of-notation for clarity (i.e. $j=\sqrt{-1}$ and $i$ is indexical). $\endgroup$ – Arturo don Juan Mar 20 '16 at 18:58

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