Cfrgtkky

I am currently learning about generating functions and
I found an interesting one for harmonic series, $dfrac{log(1-x)}{x-1}$.

Is there any hope I could get a formula for $n$th coefficient out of this? The $n$th derivative looks messy...

In absence of formula, can I at least get some asymptotic information, like that harmonic series diverges? (Can be shown more simply, I know.)

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With the identity
$ds{pars{1 - x}^{m} = sum_{k = 0}^{infty}
{m choose k}pars{-x}^{k} =
sum_{k = 0}^{infty}{k - m - 1 choose k}x^{k}}$:

It is useful to notice that multiplication of a series $A(x)=sum_{n=0}^infty a_nx^n$ with the geometric series $frac{1}{1-x}$ transforms the sequence $(a_n)_{ngeq 0}$ to a sequence of sums $left(sum_{k=0}^na_kright)_{ngeq 0}$. We obtain

begin{align*}
frac{1}{1-x}A(x)&=frac{1}{1-x}sum_{n=0}^infty a_n x^n\
&=sum_{n=0}^infty left(sum_{k=0}^n a_kright) x^ntag{1}
end{align*}

So, it is sufficient to determine the $n$-th coefficient of $A(x)$ in order to also know the $n$-th coefficient of $frac{1}{1-x}A(x)$.

Since the series expansion of $-log (1-x)$ is known to be
begin{align*}
-log(1-x)=sum_{n=1}^infty frac{x^n}{n}qquadqquadqquad |x|<1
end{align*}
it follows from (1)
begin{align*}
-frac{log(1-x)}{1-x}&=frac{1}{1-x}sum_{n=1}^infty frac{x^n}{n}
=sum_{n=1}^infty left(sum_{k=1}^nfrac{1}{k}right)x^n\
&=sum_{n=1}^infty H_nx^n
end{align*}