Bernoulli numbers.

Question

Today I read a short article about Bernoulli numbers. This article give the way to calculate Bernoulli numbers.

$\displaystyle \frac{x}{e^{x}-1} = f(x)$

$\displaystyle f(x)(e^{x}-1) = x = \sum_{n = 0}^{\infty}B_{n} \frac{x^{n}}{n!} \sum_{k = 1}^{\infty} \frac{x^{k}}{k!} = \sum_{n=1}^{\infty}x^{n} \sum_{k+m = n}^{\infty}\frac{B_{m}}{k!m!} = \sum_{n=1}^{\infty} \frac{x^{n}}{n!} \sum_{m=0}^{n-1} \binom{n}{m} B_{m}$ Now we could calculate Bernoulli numbers for different $n$. But I don't understand the step with $k+m = n$. Please help me with it!

Answer 1

In slow motion $$ \begin{array}{l} \sum\limits_{0\, \le \,n} {B_{\,n} \frac{{x^{\,n} }}{{n!}}} \sum\limits_{1\, \le \,k} {\frac{{x^{\,k} }}{{k!}}} = \sum\limits_{\begin{array}{*{20}c} {0\, \le \,n} \\ {1\, \le \,k} \\ \end{array}} {B_{\,n} \frac{{x^{\,n + k} }}{{n!k!}}} = \sum\limits_{\begin{array}{*{20}c} {0\, \le \,n} \\ {n + 1\, \le \,n + k} \\ \end{array}} {B_{\,n} \frac{{x^{\,n + k} }}{{n!k!}}} = \\ = \sum\limits_{\begin{array}{*{20}c} {0\, \le \,n} \\ {n + 1\, \le \,q} \\ \end{array}} {B_{\,n} \frac{{x^{\,q} }}{{n!\left( {q - n} \right)!}}} = \sum\limits_{\begin{array}{*{20}c} {1\, \le \,q} \\ {0\, \le \,n\, \le \,q - 1} \\ \end{array}} {B_{\,n} \frac{{x^{\,q} }}{{n!\left( {q - n} \right)!}}} = \\ = \sum\limits_{\begin{array}{*{20}c} {1\, \le \,q} \\ {0\, \le \,n\, \le \,q - 1} \\ \end{array}} {B_{\,n} \frac{{q!}}{{n!\left( {q - n} \right)!}}\frac{{x^{\,q} }}{{q!}}} = \sum\limits_{1\, \le \,q} {\left( {\sum\limits_{0\, \le \,n\, \le \,q - 1} {B_{\,n} \left( \begin{array}{l} q \\ n \\ \end{array} \right)} } \right)\frac{{x^{\,q} }}{{q!}}} = \\ = \sum\limits_{1\, \le \,n} {\left( {\sum\limits_{0\, \le \,m\, \le \,n - 1} {B_{\,m} \left( \begin{array}{l} n \\ m \\ \end{array} \right)} } \right)\frac{{x^{\,n} }}{{n!}}} \\ \end{array} $$

Answer 2

It's not your fault at all, the notation in your citation is very misleading for two reasons (see below).

This step has to do with converting the double sum

$$\sum_{n=0}^\infty \sum_{k=1}^\infty$$

to a summation over different variables. Namely, $l = m + k$ and $k$. The bounds for $l$ are $1$ to $\infty$ and for $k$, $1$ through $l$. So the sum after this step can be written in an equivalent form,

$$\sum_{n=0}^\infty \sum_{k=1}^\infty B_n \frac{x^{n+k}}{n!k!} = \sum_{l=1}^\infty \sum_{k=1}^l B_{l-k}\frac{x^l}{k!(l-k)!}.$$

Now seeing that both $k$ and $l-k$ are used in the summand, instead of explicitly iterating over one variable or the other we can write that as iterating over partitions of $l$ into two variables, $k$ and $m$:

$$\sum_{l=1}^\infty \sum_{k=1}^l B_{l-k}\frac{x^l}{k!(l-k)!} = \sum_{l=1}^\infty \sum_{\scriptstyle k \ge 1, m \ge 0 \atop \scriptstyle k+m = l} B_m\frac{x^l}{k!m!}.$$

Now you're there, except that

1. the authors just reused the name $n$ for my $l$, which is a different variable (actually, $m$ is the same as $n$ was!),

2. the infinity upper bound above $\sum_{k+m=n}$ makes no sense.

From there on, it's easier: first, we just decide that in the inner sum we might actually like iterating over one of the variables, after all, but not $k$ but $m$ this time, which (as $1 \le k \le l$) goes from $0$ to $l-1$. Expressing $k$ as $l-m$, we get

$$\sum_{l=1}^\infty \sum_{m=0}^{l-1} B_m\frac{x^l}{m!(l-m)!}.$$

Second, we notice that the division by $m!$ and $(l-m)!$ looks very much like a binomial coefficient, just except the factorial $l!$ in the numerator, so we extend by $1 = l!/l!$ and rewrite

$$\sum_{l=1}^\infty \sum_{m=0}^{l-1} B_m\frac{x^l\cdot l!}{l!\cdot m!(l-m)!} = \sum_{l=1}^\infty \sum_{m=0}^{l-1} B_m \frac{x^l}{l!} \binom{l}{m}.$$

Third, we take all we can out of the inner sum, i.e., all that does not depend on $m$.