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In a problem that I am trying to solve using generating function, the right-hand side (RHS) of the generating function equation is

$\sum_{n=1}^{\infty} (n-1)!x^n$.

Would like to find the closed form for it.

Appreciating any help.

Haigeng W
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    That series diverges for $x\ne 0$. – Mark Viola May 26 '20 at 20:58
  • Thanks. A bit correction to my original problem statement, it's ok that the solution contains some well known series (e.g. Harmonic), and an approach to finding the exact solution would be appreciated. Also, since this is the RHS of a generating function, it's ok that series is not convergent. The page at the given link (thanks, Jean) indeed discussed a similar problem but no exact solution was given there. – Haigeng W May 26 '20 at 23:51

1 Answers1

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If you rewrite the coefficient $$ (k-1)! = k^{k-1}(1-\frac{1}{k})^{k-1} \geq k^{k-1} \frac{1}{k} = k^{k-2} $$ the penultimate step is by Bernoulli inequality, since $-\frac{1}{k} > -1 \ \forall \ k>1$ So your summand is lower-bounded by $$ x^2 \sum_{k=2}^{\infty} (k x)^{k-2} $$ You can find for which $x$ it doesn't tend to $0$, so the sequence diverges, and by comaprison test so does the original sequence.

Alex
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  • Thanks. A bit correction to my original problem statement, it's ok that the solution contains some well known series (e.g. Harmonic), and an approach to finding the exact solution would be appreciated. Also, since this is the RHS of a generating function, it's ok that series is not convergent. – Haigeng W May 26 '20 at 23:52