Does $\sum_{m=0}^{\infty} {2m\choose m} \frac{1}{4^{m}} $ converge?

Solution 1:

Use Striling's approximation to see that $\binom {2m} {m}4^{-m} \geq c m^{-1/2}$ for some $c>0$. Hence the series is divergent.

Solution 2:

Here is an elementary ''non-Stirling'' approach:

\begin{eqnarray*} \frac{1}{4^m}\binom{2m}{m} & = & \frac{1}{4^m}\cdot \frac{\prod_{i=1}^m 2i \cdot \prod_{i=1}^m (2i-1) }{(m!)^2} \\ & = & \frac{1}{4^m}\cdot 4^m\frac{\prod_{i=1}^m i \cdot \prod_{i=1}^m \left(i-\frac{1}{2}\right) }{\prod_{i=1}^m i \cdot \prod_{i=1}^m i} \\ & = & \prod_{i=1}^m \left( 1 - \frac{1}{2i} \right)\\ & = & \prod^m_{i=\color{blue}{1}} \frac{2i-1}{2i} \\ & \color{blue}{>} & \frac{1}{2} \prod^m_{i=\color{blue}{2}} \frac{2i-2}{2i} \\ & = & \frac{1}{2} \prod_{i=2}^m \frac{i-1}{i} \\ & = & \frac{1}{2m} \\ \end{eqnarray*}

Hence, the given series has the divergent minorant $\sum_{m=1}^{\infty}\frac{1}{2m}$.