This question is from Ponnusamy and silvermann complex analysis Pg 436 .
Question : Suppose that $0\leq |a_1|\leq |a_2| \leq |a_3| \ldots \to \infty$. Show that $\prod_{n=1}^{\infty} ( 1- z/a_n) e^{Q_n(z) }$ represents an entire function with $Q_n(z) = z/a_n + (z/a_n)^2/2 + \ldots + (z/a_n)^{[\ln n]}/[\ln n]$ .
I attempted the question on the same lines as I attempted Show that This infinite product is entire
When in last step I have to use Weierstrass Theorem, I got the series ${1/a_n}^{[\ln n]+1}$. This series is to be proved convergent. But I am unable to prove it. I am uncertain on which result should I use.
Please help with it. Rest of details of solutions I checked and They are correct.
Best Answer
The product is $\prod_{n=1}^\infty E_{\lfloor \ln(n) \rfloor}\left( \frac{z}{a_n}\right) $ where $E_n$ are the so-called elementary factors: $$ E_{n}(z)=\begin{cases} (1-z)&{\text{if }}n=0 \, ,\\ (1-z)\exp \left({\frac {z^{1}}{1}}+{\frac {z^{2}}{2}}+\cdots +{\frac {z^{n}}{n}}\right)&{\text{otherwise}} \, . \end{cases} $$ The elementary factors satisfy the inequality $|1-E_n(z)| \le |z|^{n+1}$ for $|z| \le 1$.
Now fix $R > 0$ and choose an index $N$ such that $|a_n| > 3R$ for $n \ge N$. Then $$ \left| 1 - E_{\lfloor \ln(n) \rfloor}\left( \frac{z}{a_n}\right) \right| \le \left( \frac 13 \right)^{\lfloor \ln(n) \rfloor + 1} \le \frac{1}{3^{\ln n}} = \frac{1}{n^{\ln 3}} $$ for $n \ge N$ and $|z| \le R$.
Using the Weierstrass M-test (note that $\ln 3 > 1$) it follows that $\sum_{n=1}^\infty \left( 1 - E_{\lfloor \ln(n) \rfloor}\left( \frac{z}{a_n}\right) \right)$ converges absolutely and uniformly on every disk $|z| \le R$, and that implies that $\prod_{n=1}^\infty E_{\lfloor \ln(n) \rfloor}\left( \frac{z}{a_n}\right) $ converges to a holomorphic function on $\Bbb C$, with zeros exactly at the $a_n$.