In Apostol Mathematical Analysis Exercise 10.31, the question want us to show that $$\Gamma(x)=\sum_{n=0}^\infty \dfrac{(-1)^n}{n!}\dfrac1{n+x}+\sum_{n=0}^\infty c_nx^n$$ for $x>0$ where $c_n=(1/n!)\int_1^\infty t^{-1}e^{-t}(\log t)^ndt$. This is an easy one. The question after it is asking to show that the complex series $$\sum_{n=0}^\infty c_nz^n$$ converges for $z\in \mathbb C$. I find this very difficult for me to prove, maybe I'm missing something.

My approach is using the ratio test $$\left|\dfrac{c_{n+1}z^{n+1}}{c_nz^n}\right|=\dfrac{c_{n+1}}{c_n}|z|<1$$ so we need to show that $$0=\lim_{n\to\infty}\dfrac{c_{n+1}}{c_n}=\lim_{n\to\infty}\dfrac1{n+1}\dfrac{\int_1^\infty t^{-1}e^{-t}(\log t)^{n+1}dt}{\int_1^\infty t^{-1}e^{-t}(\log t)^ndt}$$ so the ratio test will always return $<1$ for every $z\in\mathbb C$.

Notice the integrands in both integrals, I've come up an idea which let $f(t)=t^{-1}e^{-t}(\log t)^n, g(t)=\log t$, then I use the Intermediate value theorem for integrals $$\int_1^\infty f(t)g(t)dt=g(c)\int_1^\infty f(t)dt$$ for some $c>1$. For this I'm trying first not to consider that this integral is improper, then find out the value of $c$, or some reasonable bound of $c$, but I'm stuck from here. My expectation is that the ratio of integrals is of order $\log n$, but logically speaking it is good enough if the ratio is $o(n)$.

Another idea of mine is to show $$0=\lim_{n\to\infty}\sqrt[n]{c_n}=\lim_{n\to\infty}\sqrt[n]{\frac1{n!}\int_1^\infty t^{-1}e^{-t}(\log t)^ndt},$$ but this seems even harder.

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    $\begingroup$ Hint: if the series converges for all postive numbers $z$ then it converges for all complex $z$. For positive $z$ interchange the sum and the integral. $\endgroup$ Aug 9, 2020 at 8:13
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    $\begingroup$ $\sum_{n=0}^\infty c_n z^n=\int_1^\infty t^{z-1}e^{-t}\,dt$ which is an entire function of $z$. $\endgroup$
    – metamorphy
    Aug 9, 2020 at 8:14
  • $\begingroup$ Thanks, I didn't know the solution would be this elegant! $\endgroup$ Aug 9, 2020 at 8:20

1 Answer 1


This is what I understand from @KaviRamaMurthy 's comment. Since $\sum_{n=0}^\infty c_nx^n$ is the real Taylor series for the integral $$\int_1^\infty t^{x-1}e^{-t}dt$$ and this integral converges for all positive $x$, so the Taylor series also converges for all positive $x$. Now we choose any $z\in\mathbb C$ and realize that $$\sum_{n=0}^\infty |c_nz^n|\leq \sum_{n=0}^\infty c_n|z|^n,$$ which indicates the complex series is absolutely convergent for all $z$, thereby confirm the complex series converges everywhere.


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