Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

How is the Bernoulli numbers? For example, as against $B_2$? For example, found that in internet $$\sum_{n=0}^{\infty} \frac{B_n\;x^n}{n!}=\frac{x}{e^x-1}$$ but if I want to find $B_2$ then $$\sum_{n=0}^{\infty}\frac{B_2~x^2}{2}=\frac{x}{e^x-1}$$and I believe this is not much help. I want to learn how to calculate Bernoulli numbers to learn how to calculate $\zeta(2n)$.

share|cite|improve this question
Expand $\frac{e^{-x}}{1-e^{-x}}$ via geometric series then expand $e^{-nx}$s via the power series for $\exp$, or just use Taylor series directly on $x/(e^x-1)$. If you write something like $\sum B_2x^2/2$, it probably says you are not very familiar with power series. – anon Jan 18 '14 at 13:43
@anon Could you show me how to $B_2$? – marcelolpjunior Jan 18 '14 at 13:51
$B_2=\dfrac{1}{6}$ – Leox Jan 18 '14 at 13:52
@marcelolpjunior If you weren't able to follow my comment, I recommend reading up on power series and Taylor series specifically. – anon Jan 18 '14 at 13:54
If you simply want to compute $\zeta(2n)$, you can use this recursion. – robjohn Jan 18 '14 at 16:22

Note that $$\frac{e^z-1}z=\frac1 z\sum_{n=1}^\infty\frac1{n!}z^n=\sum_{n=1}^\infty\frac1{n!}z^{n-1}=\sum_{n=0}^\infty\frac1{(n+1)!}z^n$$ and we can use Mertens’ multiplication theorem to get $$1=\left(\sum_{n=0}^\infty\frac{B_n}{n!}z^n\right)\left(\sum_{n=0}^\infty\frac1{(n+1)!}z^n\right)=\sum_{n=0}^\infty\sum_{k=0}^n\left(\frac{B_k}{k!}\frac{1}{(n-k+1)!}\right)z^n$$ By the identity theorem, the $n=0$ term on the right must equal $1$ while all other terms must vanish. The $n=0$ term on the right is just $B_0$, so $B_0 = 1$, and for $n > 1$, we must have $\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}=0$. Multiplying this by $(n + 1)!$ we get $$0=\sum_{k=0}^n\frac{B_k}{k!}\frac{1}{(n-k+1)!}=\sum_{k=0}^n\frac{(n+1)!}{k!(n-k+1)!}B_k=\sum_{k=0}^n\binom{n+1}kB_k$$ and adding $B_{n+1}=\binom{n+1}{n+1}B_{n+1}$ to both sides of this equation, we get $$B_{n+1}=\sum_{k=0}^{n+1}\binom{n+1}kB_k$$ The right-hand side might look familiar from the binomial formula. Recall from the binomial formula that for any complex number $a$, we have $$(a+1)^{n+1}=\sum_{k=0}^{n+1}\binom{n+1}ka^k1^{n+1-k}=\sum_{k=0}^{n+1}\binom{n+1}ka^k$$ Notice that the right-hand side of this expression is exactly the right-hand side of the previous equation if put $a = B$ and we make the superscript $k$ into a subscript $k$. Thus, if we use the notation $\Doteq$ to mean equals after making superscripts into subscripts, then we can write $$\boxed{B^{n+1}\Doteq (B+1)^{n+1},n=1,2,3,...,B_0=1}$$ Use recent identity, one can in principle find all the Bernoulli numbers: When $n = 1$, we see that $$B^2\Doteq(B+1)^2=B^2+2B^1+1\Rightarrow0=2B_1+1\Rightarrow B_1=\frac{-1}2.$$ When $n = 2$, we see that $$B^3\Doteq(B+1)^3=B^3+3B^2+3B^1+1\Rightarrow3B_2+3B_1+1=0\Rightarrow B_2=\frac1 6.$$

share|cite|improve this answer
This has a very similar feel to this answer, which develops a recursion for $\zeta(2n)$. (+1) – robjohn Jan 18 '14 at 16:29

As suggested by anon, develop $\dfrac{x }{e^x - 1}$ as a Taylor series. You then obtain for the rhs

$$1 - \frac x 2 + \frac{x^2}{12} - \frac{x^4}{ 720} + \frac{x^6}{ 30240} - \frac{x^8}{ 1209600} + \cdots$$

The lhs write

$$B_0 + B_1 x + B_2\frac{x^2}{2!} + B_3 \frac{x^3}{3!} + B_4\frac{x^4}{4!} +\cdots$$

Then $$B_0 = 1, B_1 = -1/2, B_2 = 1/6, B_3 = 0, B_4 = -1/30 ....$$

share|cite|improve this answer
@Pedro Tamaroff. Thanks for editing for me. Cheers. – Claude Leibovici Jan 18 '14 at 14:39
No problem. ${}{}$ – Pedro Tamaroff Jan 18 '14 at 14:41

There is an explicit formula $$ B_m = \sum_{k=0}^m\sum_{v=0}^k(-1)^v\binom kv\frac{ v ^m}{k+1} $$

Edit. Put $P(k,v)=\displaystyle (-1)^v\binom kv\frac{ v ^m}{k+1}.$

Then $B_2=P \left( 0,0 \right) +P \left( 1,0 \right) +P \left( 1,1 \right) +P \left( 2,0 \right) +P \left( 2,1 \right) +P \left( 2,2 \right).$

Taking into account the calculation ${\it P} \left( 0,0 \right) =0,{\it P} \left( 1,0 \right) =0,{\it P } \left( 1,1 \right) =-1/2,{\it P} \left( 2,0 \right) =0,{\it P} \left( 2,1 \right) =-2/3,{\it P} \left( 2,2 \right) =4/3 $ you will get easilly $B_2.$

share|cite|improve this answer
Who are $v$ and $k$? $$B_2=\sum_{k=0}^{2}\sum_{v=0}^{k}(-1)^v\binom kv \frac{v^2}{k+1}$$ – marcelolpjunior Jan 18 '14 at 14:03
That are indexes of summation – Leox Jan 18 '14 at 14:04
Make this the two for me, so I try to do next? I find it easier to understand as well. Thank you! – marcelolpjunior Jan 18 '14 at 14:07
@marcelolpjunior Your above comment on this answer tells you exactly what you would need to do next. If you are not familiar with summation notation, you will have to look that up and become familiar with it. If you came to this problem of finding Bernoulli numbers without even knowing about power series or summation notation or derivatives and whatnot, you do not yet have the tools to study Bernoulli numbers. – anon Jan 19 '14 at 6:05

Compare $$\sum_{n=0}^{\infty} \frac{B_nx^n}{n!}=B_0+B_1x+\frac{B_2x^2}2+\frac{B_3x^3}6+\cdots\left(=\frac{x}{\mathrm e^x-1}\right) $$ with $$ \sum_{n=0}^{\infty} \frac{B_2x^n}{2}=\frac{B_2}2+\frac{B_2x}2+\frac{B_2x^2}2+\frac{B_2x^3}2+\cdots\left(=\frac{B_2}{2(1-x)}\right)$$ To find $B_2$, either compute $f''(0)$ for $f:x\mapsto\dfrac{x}{\mathrm e^x-1}$, or expand $f(x)$ into powers of $x$ up to the power $x^2$ (this will involve expanding $\mathrm e^x$ into powers of $x$ up to the power $x^3$, but not more). And indeed, the result is $B_2=\dfrac16$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.