TAOCP 4.5.2 Exercise 27

Let $\psi_n$ be defined by the logarithmic generating function that precedes (58): \log\frac{x}{e^x-1}=\sum_{n\ge1}\psi_n x^n .

taocpmathematicsalgorithmsvolume-2

Section 4.5.2: The Greatest Common Divisor

Exercise 27. [**] [M22] Prove (58), which expresses $\psi_n$ in terms of Bernoulli numbers.

Verified: yes
Solve time: 1m24s

Let $\psi_n$ be defined by the logarithmic generating function that precedes (58):

$$ \log\frac{x}{e^x-1}=\sum_{n\ge1}\psi_n x^n . $$

Differentiate:

$$ \sum_{n\ge1}n\psi_n x^{,n-1} = \frac1x-\frac{e^x}{e^x-1}. $$

Since the Bernoulli numbers are defined by

$$ \frac{x}{e^x-1} = \sum_{m\ge0}\frac{B_m}{m!}x^m, $$

we have

$$ \frac{e^x}{e^x-1} = \frac1x+\sum_{m\ge1}\frac{B_m}{m!}x^{m-1}. $$

Hence

$$ \frac1x-\frac{e^x}{e^x-1} = -\sum_{m\ge1}\frac{B_m}{m!}x^{m-1}. $$

Comparing coefficients of $x^{n-1}$ gives

$$ n\psi_n=-\frac{B_n}{n!}, $$

therefore

$$ \psi_n=-\frac{B_n}{n,n!}. $$

This is exactly formula (58), expressing $\psi_n$ in terms of the Bernoulli numbers. Since $B_{2m+1}=0$ for $m\ge1$, (58) also implies that $\psi_{2m+1}=0$ for $m\ge1$. ∎