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Show that the series defines and entire function, $$\sum_{n=1}^\infty {{z(z+1)\cdots (z+n-1)}\over n^n}.$$

I that that an entire function is one that is analytic at every point in the plane. This series looks like it is an infinite sum of polynomials where each polynomial is entire. I'm not really sure how to expand this because of the numerator or how to simplify it so that I can see how it is an entire function. Any solutions or hints are greatly appreciated.

Happy
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1 Answers1

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Remember the following criterion

If $f_n$ is a sequence of entire functions such that $\sum_{n=1}^{+\infty} f_n$ is uniformly convergent on each compact, then $\sum_{n=1}^{+\infty} f_n$ is entire.

So, the only think you have to do is to prove the uniform-convergence of your series. For example you can prove that the series is Cauchy uniform on each disk $D(0,r)$ (It is clear that every compact is included in such a disk). We have $$\sup_{|z|<r}\left|\sum_{n=p}^q {{z(z+1)\cdots (z+n-1)}\over n^n}\right| \leq \sum_{n=p}^q{{r(r+1)\cdots (r+n-1)}\over n^n} \to 0$$ if $p,q \to +\infty,$ since $$\sum_{n=1}^{+\infty}{{r(r+1)\cdots (r+n-1)}\over n^n}$$ is absolutely convergent by the quotient criterion.

C. Dubussy
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  • Thank you! I haven't seen this criterion before. Is it in any standard complex textbooks? Is it common? – Happy Apr 24 '16 at 00:01
  • Yes, it is normaly one of the first criterion we see in complex analysis. It is often called the Weierstrass theorem. See here for example : https://www.encyclopediaofmath.org/index.php/Weierstrass_theorem#Uniformly_convergent_series_of_analytic_functions – C. Dubussy Apr 24 '16 at 00:06