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

Question

I would like to check if the series $\sum_{m=0}^{\infty} {2m\choose m} \frac{1}{4^{m}} $ converges or diverges. The ratio test is inconclusive. Any other way to proceed? Thanks.

Answer 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.

Answer 2

$$ \begin{align} \frac1{4^n}\binom{2n}{n} &=\frac{2n(2n-1)}{4n^2}\frac1{4^{n-1}}\binom{2n-2}{n-1}\\ &=\frac{n-\frac12}{n}\frac1{4^{n-1}}\binom{2n-2}{n-1}\\ \end{align} $$ Thus, $$ \frac1{4^n}\binom{2n}{n}=\prod_{k=1}^n\frac{k-\frac12}{k} $$ and therefore, $$ \begin{align} \left[\frac1{4^n}\binom{2n}{n}\right]^2 &=\prod_{k=1}^n\frac{k^2-k+\frac14}{k^2}\\ &\ge\frac14\prod_{k=2}^n\frac{k-1}{k}\\ &=\frac1{4n} \end{align} $$ So we see that for $n\ge1$, $$ \frac1{4^n}\binom{2n}{n}\ge\frac1{2\sqrt{n}} $$ so the sum diverges.

Answer 3

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 $\sum_{m=1}^{\infty}\frac{1}{2m}$.

Answer 4

According to the Sterling approximation, $$n! \approx \sqrt {2\pi n}(n/e)^n$$

Therefore $$ {2m\choose m} \approx \frac {4^m}{\sqrt {2\pi m}}$$

Which makes the series $\sum_{m=0}^{\infty} {2m\choose m} \frac{1}{4^{m}}$diverge.

Answer 5

Considering the partial sums $$S_p=\sum_{m=0}^{p} {2m\choose m} \frac{1}{4^{m}}=\binom{p+\frac{1}{2}}{p}$$ and using Stirling approximation, we have $$\log(S_p)=\log \left(\frac{2 \sqrt{p}}{\sqrt{\pi }}\right)+\frac{3}{8 p}+O\left(\frac{1}{p^2}\right)$$ Then, continuing with Taylor series $$S_p=e^{\log(S_p)}=\frac{2 \sqrt{p}}{\sqrt{\pi }}+\frac{3}{4 \sqrt{p\pi }} +O\left(\frac{1}{p^{3/2}}\right)$$ For $S_{10}$, the exact value is $\frac{969969}{262144}\approx 3.70014$ while the truncated series gives $\frac{83}{4 \sqrt{10 \pi }}\approx 3.70206$.

For $p > 23$, the relative error is $< 0.01$%.

Answer 6

The mere divergence of $\displaystyle\sum_{m=0}^\infty \binom{2m}m \frac1{4^m}$ follows from the easy bound $$ 4^m = \sum_{k=0}^{2m} \binom{2m}k \le \sum_{k=0}^{2m} \binom{2m}m = (2m+1)\binom{2m}m, $$ so that $\displaystyle\sum_{m=0}^\infty \binom{2m}m \frac1{4^m} \ge \sum_{m=0}^\infty \frac1{2m+1}$.