# A series convergence problem about Gamma function

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.

• 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. Aug 9, 2020 at 8:13
• $\sum_{n=0}^\infty c_n z^n=\int_1^\infty t^{z-1}e^{-t}\,dt$ which is an entire function of $z$. Aug 9, 2020 at 8:14
• Thanks, I didn't know the solution would be this elegant! Aug 9, 2020 at 8:20

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.