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 $$\sum_{n=0}^\infty\frac{2n}{8^n} \binom{2n}n=\frac{1}{4}(1-1/4)^{-3/2}.$$

Answer 2

We'll use $|w|<1\implies\sum_{n\ge0}nw^n=\frac{w}{(1-w)^2}$. Your sum is$$\begin{align}\sum_{n\ge0}2n(1/8)^n\oint_{|z|=1}\frac{(1+z)^{2n}dz}{2\pi iz^{n+1}}&=\oint_{|z|=1}\frac{dz}{\pi iz}\left.\frac{w}{(1-w)^2}\right|_{(1+z)^2/(8z)}\\&=\oint_{|z|=1}\frac{dz}{2\pi i}\frac{16(1+z)^2}{(1-6z+z^2)^2}.\end{align}$$We can evaluate this with the residue theorem; the only relevant pole is $z=3-2\sqrt{2}$ but it's second order, so the sum is$$\left.\frac{d}{dz}\frac{16(1+z)^2}{(z-3-2\sqrt{2})^2}\right|_{3-2\sqrt{2}},$$which I'll show leave you to show is $\sqrt{2}$. In fact, it can be shown that$$|x|<\frac14\implies\sum_{n\ge0}2n\binom{2n}{n}x^n=4x(1-4x)^{-3/2}.$$

Answer 3

Hint:

$$ \frac{1}{\sqrt{1-4x}} = \sum_{n=0}^\infty {2n \choose n} x^n$$

is a known Taylor series which shows up all the time in relativity.

Answer 4

Hint By the Binomial series $$(1+x)^{-\frac{1}{2}}=\sum_{n=0}^\infty {{-\frac{1}{2}} \choose{n}}x^n$$

Now, $$ {{-\frac{1}{2}} \choose{n}}=\frac{(-\frac{1}{2})(-\frac{3}{2})..(-\frac{2n-1}{2})}{n!}=\frac{(-1)^n}{4^n}\frac{2n!}{n! n!}$$

Now, derivate and plug in $x=-\frac{1}{2}$.