Convergence of the complex power series $\sum_{n=1}^\infty \dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}z^n$

Question

I have to study the convergence of

$$S(z)=\displaystyle \sum_{n=1}^\infty \dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}z^n,\quad \quad z\in \mathbb{C}$$

I have defined $c_n=\dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}$ and I calculated, first of all, the convergence radius $R$ of the series. I have observed that

$$\dfrac{|c_{n+1}|}{|c_n|}=\dfrac{2n+1}{2n+2}\rightarrow 1,\quad n\to \infty,$$

Therefore, $R=1$ and the convergence theorem of the power series ensures that

• $S(z)$ converges absolutely (and punctually) for all $z\in \mathbb{D}(0,1)$.
• $S(z)$ converges uniformly on all $\mathbb{D}(0,1)$-compacts.
• $S(z)$ diverge for all $z$ with $|z|>1$.

Now I was trying to see what happens to the $\mathbb{D}(0,1)$-frontier points. Let $z$ with $|z|=1$ and $b_n$ be the general term for $S(z)$. So,

$$|b_n|=\dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}|z|^n=\dfrac{1\cdot 3\cdot 5\cdots (2n-1)}{2\cdot 4\cdots (2n)}=\dfrac{\frac{(2n)!}{2^n\cdot n!}}{2^n\cdot n!}=\dfrac{(2n)!}{4^n\cdot (n!)^2}\sim \dfrac{\sqrt{4\pi n}}{2\pi n}\to 0,\quad n\to \infty$$ applying Stirling. I don't know how to continue the study of the series on the border. I was trying with $z=e^{i\theta}$ and the criteria of Dirichlet and Abel, but I can't get anywhere.

Answer 1

Actually, Dirichlet's test works when $z\neq1$. We have $c_n$ decreasing to $0$, and if we take $b_n=z^n$, then $$\left|\sum_{n=1}^Mb_n\right|=\left|\sum_{n=1}^Mz^n\right|=\left|\frac{1-z^{M+1}}{1-z}\right|\leq\frac2{|1-z|}$$ so the series converges at $z$.

The series fails to converge at $z=1$ by Gauss's test:

$$\left|\frac{c_n}{c_{n+1}}\right|=\frac{2n+2}{2n+1} = 1+\frac1{2n+1}=1+\frac1{2n}-\frac1{2n(2n+1)}$$ Now we can take $$h=\frac12,\ r=2,\ B(n)=\frac{-n^2}{2n(2n+1)}$$ and since $h<1$, the series diverges.

Answer 2

Using $$ \frac{{1 \cdot 3 \cdots (2n - 1)}}{{2 \cdot 4 \cdots 2n}} = \frac{2}{\pi }\int_0^{\pi /2} {\sin ^{2n} tdt} $$ and summation together with analytic continuation, we obtain $$ \left( {\frac{1}{{\sqrt {1 - z} }} -1= } \right)\sum\limits_{n =1}^{N - 1} {\frac{{1 \cdot 3 \cdots (2n - 1)}}{{2 \cdot 4 \cdots 2n}}z^n } + z^N \frac{2}{\pi }\int_0^{\pi /2} {\frac{{\sin ^{2N} t}}{{1 - z\sin ^2 t}}dt} $$ for all $z \in \mathbb{C}\setminus \left[ {1, + \infty } \right)$ and $N\geq 1$. Note that for $0<s<1$, $$ \left| 1 - zs \right|^2 = \left| z \right|^2 s^2 - 2(\Re z)s + 1 \ge \begin{cases} 1 & \text{ if } \; \Re z \le 0, \\ \frac{(\Im z)^2}{ \left| z \right|^2} & \text{ if } \; 0<\Re z \le \left| z \right|^2, \\ \left|1-z\right|^2 & \text{ if } \; \Re z > \left| z \right|^2.\end{cases} $$ Consequently, the remainder term in absolute value is at most $$ \left| {\frac{{1 \cdot 3 \cdots (2N - 1)}}{{2 \cdot 4 \cdots 2N}}z^N } \right| \times \begin{cases} 1 & \text{ if } \; \Re z \le 0, \\ \left|\frac{z}{\Im z}\right| & \text{ if } \; 0<\Re z \le \left| z \right|^2, \\ \frac{1}{|1-z|} & \text{ if } \; \Re z > \left| z \right|^2.\end{cases} $$ On the unit circle, $$ \left| {\frac{{1 \cdot 3 \cdots (2N - 1)}}{{2 \cdot 4 \cdots 2N}}z^N } \right| < \frac{1}{{\sqrt {\pi N} }} $$ by the known inequality for the central binomials. Hence, on the unit circle $$ \left| {\sum\limits_{n = N}^\infty {\frac{{1 \cdot 3 \cdots (2n - 1)}}{{2 \cdot 4 \cdots 2n}}z^n } } \right| \le \frac{1}{{\sqrt {\pi N} }} \times \begin{cases} 1 & \text{ if } \; \Re z \le 0,\,|z|=1, \\ \frac{1}{|\Im z|} & \text{ if } \; 0<\Re z \le 1,\,|z|=1.\end{cases} $$ This shows that the series is convergent at every point of the circle except perhaps at $z=1$. Clearly, it cannot converge at $z=1$.