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Michael Spivak in Calculus, Fourth ed., Chapter 27, Problem 16 (page 572) defines the Bernoulli numbers based on $$ \frac{z}{e^z-1} = \sum_{n=0}^\infty\frac{B_nz^n}{n!}. $$ He asks the reader to prove that $B_n = 0$ if $n$ is odd and $> 1$. This is easy with the suggestion that Spivak provides. In a footnote he remarks that the numbers $B_{2n}$ alternate in sign but says "we will not prove this." I didn't see how to do it and embarked on a web search to find a relatively simple proof but failed. Can anyone point me to an easy proof?

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4 Answers 4

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Apparently the Bernoulli numbers can be defined in terms of the Riemann-zeta function;

$$B_n=-n\zeta(n-1)$$

An application of the zeta functional equation and the gamma reflection formula yields;

$$B_{2n}={{(-1)^{n+1}\cdot n \cdot (2n)!}\over{(2\pi)^{2n}}}\zeta(2n)$$

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The hint below is key to the following argument.

I should have clarified that I wanted to use the definition of Bernoulli numbers based on $$ \sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac{x}{e^x-1}. $$

From this one can show that $$ \tan(x)=\sum_{n=1}^\infty(-1)^{n-1}B_{2n}a_nx^{2n-1},\tag 1 $$ where $a_n>0$ for all $n$. See Bernoulli numbers, taylor series expansion of tan x for a proof.

It is easy to prove by induction that for even $n$ the $n^{th}$ derivative of $\tan(x)$ is of the form $$ \tan^{(n)}(x)=\sum_{j=0}^{n/2}b_{n,j}\tan^{2j+1}(x) $$ and for odd $n$ of the form $$ \tan^{(n)}(x)=\sum_{j=0}^{(n+1)/2}b_{n,j}\tan^{2j}(x), $$ where $b_{i,j}>0$ for all $i$ and $j$. The proof depends on the fact that $\tan^{'}(x) = 1 + \tan^2(x)$.

The Maclaurin expansion is therefore $$ \tan(x)=\sum_{n=1}^\infty \frac {b_{(2n-1),0}}{(2n-1)!}x^{2n-1}\tag 2 $$

Equating the coefficients in (1) and (2), we see that $(-1)^{n-1}B_{2n} > 0$.

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  • $\begingroup$ +1, Great answer! $\endgroup$
    – orangeskid
    Commented Aug 23, 2022 at 1:00
  • $\begingroup$ From your results, you also get: $\tan^{(n)} >0$ for $n$ odd, and for every $x$, $\tan^{(n)} x$ have the same sign for all $n$ even, although that also follows from the series expansion post-factum $\endgroup$
    – orangeskid
    Commented Aug 24, 2022 at 20:28
  • $\begingroup$ Such is so. $\tan^{(n)}(x) > 0$ for positive $x$ near $0$. $\tan^{(n)}(0) = 0$ for even $n$ and is $> 0$ for odd $n$. Also, as for any odd function, $\tan^{(n)}(-x) = (-1)^{n-1} \tan^{(n)}(x)$. $\endgroup$ Commented Aug 25, 2022 at 21:40
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HINT: Use the definition of the Bernoulli numbers and the following fact: the function $\tan x$ has a Taylor expansion at $0$ with all coefficients $\ge 0$. This can be shown by induction using the equality

$$(\tan x)' = 1 + \tan^2 x$$ and $\tan 0 = 0$.

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    $\begingroup$ Many thanks for the hint. See my answer below. $\endgroup$ Commented Aug 23, 2022 at 0:48
  • $\begingroup$ @Richard Hevener: You are very welcome! $\endgroup$
    – orangeskid
    Commented Aug 23, 2022 at 1:01
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There is also a purely formal proof. The details are given in Corollary 1.16 of [T. Arakawa, T. Ibukiyama, M. Kaneko, Bernoulli Numbers and Zeta Functions, Springer, 2014].

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