Evaluate the sum $\sum_{n=0}^{\infty} \frac{2n}{8^n}{{2n}\choose{n}}$

Question

Evaluate the sum $\sum_{n=0}^{\infty} \frac{2n}{8^n}{{2n}\choose{n}}$

I am unable to find a way to solve this sum. I have never seen sums involving binomial coefficients multiplied by $n$. Help will be appreciated

Answer 1

Note that \begin{align}\binom{2i}{i} \frac1{2^{2i}}&=\frac{\left(\prod\limits_{j=1}^i2j\right)\left(\prod\limits_{j=0}^{i-1}(2j+1)\right)}{2^i 2^i(i!)^2}\\ &=(-1)^i\frac{(-1/2)(-1-1/2)\ldots (-1/2-i+1)}{i!}=(-1)^i\binom{-1/2}i\end{align} and therefore one can rewrite the power series (for $x\in\Bbb C$ with $|x| <1$) $$\tag{1}\sum_{n=0}^\infty \binom{2n}n \frac{x^{n}}{4^{n}}=\sum_{n=0}^\infty \binom{-1/2}n(-x)^n=(1-x)^{-1/2},$$ where we have used the Binomial series.

To establish a formula for your series, just differentiate (1) to find that $$\sum_{n=1}^\infty 2n l\binom{2n}n \frac{x^{n-1}}{4^n}=(1-x)^{-3/2}$$ and hence $$\sum_{n=0}^\infty 2n \binom{2n}n \frac{x^{n}}{4^{n}}=x(1-x)^{-3/2}.$$ Thus, taking $x= 1/4$ one gets $$\sum_{n=0}^\infty\frac{2n}{8^n} \binom{2n}n=\frac{1}{4}(1-1/4)^{-3/2}.$$