Calculating Bernoulli Numbers from $\sum\limits_{n=0}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$

Question

How is the Bernoulli numbers? For example, found that in internet $$\sum_{n=0}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$$ but if I want to find $B_2$ then $$B_0+B_1x+\frac{B_2x^2}{2}+\sum_{n=3}^\infty\frac{B_nx^n}{n!}=\frac x{e^x-1}$$ and I believe this is not much help. I want to learn how to calculate Bernoulli numbers to learn how to calculate $\zeta(2n)$.

Answer 1

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 $$B^3\Doteq(B+1)^3=B^3+3B^2+3B^1+1\Rightarrow3B_2+3B_1+1=0\Rightarrow B_2=\frac1 6.$$

Answer 2

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

Answer 3

The Bernoulli numbers are famous enough to have their own webpage. It gathers much that is known about them, and gives an efficient program to compute them (open source, under GPL). The extensive bibliography includes algorithms too.

Answer 4

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 $B_2.$

Answer 5

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 $B_2=\dfrac16$.