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

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 $B_2=\dfrac16$.

Answer 2

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 3

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 4

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