Spring 2018 - Problem 9

Solution.

1. It suffices to show that

   $$\sum_{n=1}^\infty \br{1 - \p{1 + \frac{1}{n}}^z \p{1 - \frac{z}{n}}}$$

   converges. First, by Taylor expansion about $x = 0$,

   $$\p{1 + x}^z = 1 + zx + O\p{x^2}.$$

   Thus,

   $$\begin{aligned} \br{1 - \p{1 + \frac{1}{n}}^z \p{1 - \frac{z}{n}}} &= \frac{n - \p{n - z}\p{1 + \frac{1}{n}}^z}{n} \\ &= \frac{1}{n}\p{n - \p{n - z}\p{1 + \frac{z}{n} + O\p{\frac{1}{n^2}}}} \\ &= \frac{1}{n} \p{\frac{z^2}{n} + \p{z - n}O\p{\frac{z^2}{n^2}}} \\ &= O\p{\frac{1}{n^2}}, \end{aligned}$$

   so the series converges and hence the product converges as well.

2. Let $K$ be compact. From the calculation above, we see that the corresponding sum to the product converges uniformly on $K$, so eventually all terms lie in the disk $B\p{1, \frac{1}{2}}$. If $f_n\p{z} = \p{1 + \frac{1}{n}}^z \p{1 - \frac{z}{n}}$, then

   $$\begin{gathered} \log\p{\prod_{n=N}^M f_n\p{z}} = \sum_{n=N}^M \log{f_n\p{z}} = \sum_{n=N}^M \log\p{1 + \p{f_n\p{z} - 1}} = \sum_{n=N}^M O\p{f_n\p{z} - 1} \\ \implies \prod_{n=N}^M f_n\p{z} = \exp\p{\sum_{n=N}^M O\p{\abs{f_n\p{z} - 1}}} = \exp\p{\sum_{n=N}^M O\p{\frac{1}{n^2}}}, \end{gathered}$$

   where the big-O depends on $K$. In other words, the product also converges uniformly on $K$, and so the formal product extends to an entire function.