Show series $\sum_{k=1}^{\infty} \frac{(2k-2) !\left(\frac{1}{4}\right)^{k-1}\left(1-\frac{1}{4}\right)^k}{k !(k-1) !}=1.$

Question

I have the following series, together with its result

$$\sum_{k=1}^{\infty} \frac{(2k-2) !\left(\frac{1}{4}\right)^{k-1}\left(1-\frac{1}{4}\right)^k}{k !(k-1) !}=1.$$

This can be further simplified to

$$\sum_{k=1}^{\infty} \frac{2^{-4 k} \times 3^k\left(\begin{array}{c} 2 k \\ k \end{array}\right)}{2 k-1}=\frac{1}{2}.$$

It is simple enough to verify that the series is (absolutely) convergent (ratio test), however, I have no idea how to prove the equality. Any ideas for a solution? Any tips for what could usually come in useful in this case?

Answer 1

$$S=\sum_{k=1}^{\infty} \frac{(2k-2) !\left(\frac{1}{4}\right)^{k-1}\left(1-\frac{1}{4}\right)^k}{k !(k-1) !}=4\sum_{k=1}^{\infty} \frac{(2k-2) !}{k !(k-1) !} \left(\frac{3}{16}\right)^{k}$$

Define

$$f(x)=4\sum_{k=1}^{\infty} \frac{(2k-2) !}{k !(k-1) !} x^{k}\Longrightarrow f'(x)=4\sum_{k=1}^{\infty} \frac{(2k-2) !}{(k-1) !(k-1) !} x^{k-1}$$

Let $n=k-1$

$$f'(x)=4\sum_{n=0}^{\infty} \binom{2n}n x^n=\frac4{\sqrt{1-4x}}$$

Integrate and we get

$$f(x)=-2\sqrt{1-4x}+C$$

since $f(0)=0$, we get $C=2$, hence

$$f(x)=2-2\sqrt{1-4x}$$

and

$$S=f\left(\frac3{16}\right)=1$$