Does a the following Taylor series has a closed form expression \begin{align} \sum_{k=0}^\infty \frac{x^k}{k!} \log(k!). \end{align}

Note that $\Gamma(k+1)=k!$.

Also, note that this power series converges (see for example here).

  • $\begingroup$ So why not just write $\sum_{k=0}^\infty x^k(\log k!)/k!$? $\endgroup$ May 30, 2018 at 18:30
  • $\begingroup$ Now it is. Typically expression of $\log(\Gamma(z))$ have integral expression. $\endgroup$
    – Lisa
    May 30, 2018 at 18:33
  • $\begingroup$ I think you mean 'a closed form expression' since, due to the very existence of a convergent Taylor series, the corresponding function is analytic. $\endgroup$
    – rafa11111
    May 30, 2018 at 19:47
  • 1
    $\begingroup$ @rafa11111 Yes, I mean closed-form expression. I will change it. $\endgroup$
    – Lisa
    May 30, 2018 at 23:59

1 Answer 1


This is not answer, but it is a little long for a comment.

Let $$I(a)=\sum_{k=0}^{+\infty} \frac{x^k}{k!}\log \Gamma(k+a)$$ Then $$I'(a)=\sum_{k=0}^{+\infty} \frac{x^k}{k!}\psi^{(0)}(k+a)$$ And because $$\psi^{(0)}(k+a)=H_{k+a-1}-\gamma$$ We can find that \begin{align} I'(a)&=\sum_{k=0}^{+\infty} \frac{x^k}{k!}H_{k+a-1}-\sum_{k=0}^{+\infty}\gamma \frac{x^k}{k!} \\ &=\psi^{(0)}(1)e^x+\sum_{k=0}^{+\infty} \frac{x^k}{k!}H_{k+a-1} \end{align} However, I am not sure how to evaluate the last sum.

  • $\begingroup$ It seems like $I'(a)$ is not very useful for computing $I(1)$ without some other value like $I(0)$. $\endgroup$
    – Ian
    May 30, 2018 at 19:02
  • $\begingroup$ It's interesting that $I'(1)=e^x(\ln(x)+\Gamma(0,x))$. $\endgroup$
    – aleden
    May 30, 2018 at 19:45
  • $\begingroup$ The exponential generating function for the harmonic numbers may be useful. $\endgroup$
    – Plopperzz
    May 31, 2018 at 0:08
  • $\begingroup$ Found it $$\sum_{n=1}^{\infty}\frac{H_{n}}{n!}x^n = -e^{x}\left(\text{Ei}(-x)-\ln{|x|}-\gamma \right)$$ $\endgroup$
    – Plopperzz
    May 31, 2018 at 0:18

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