Problem source

\begin{questionparts}
\item  Write down the general term in the expansion
in  powers of $x$  of  $(1-x)^{-1}$,  $(1-x)^{-2}$ and   $(1-x)^{-3}$, where $|x| <1$.
Evaluate 
$\displaystyle \sum_{n=1}^\infty n 2^{-n}$ and 
$\displaystyle \sum_{n=1}^\infty n^22^{-n}$.
\item   Show that $\displaystyle (1-x)^{-\frac12} = \sum_{n=0}^\infty \frac{(2n)!}{(n!)^2}
\frac{x^n}{2^{2n}}$ , for $|x|<1$.
Evaluate $\displaystyle \sum_{n=0}^\infty \frac{(2n)!} {(n!)^2 2^{2n}3^{n}} $ and $\displaystyle \sum_{n=1}^\infty \frac{n(2n)!} {(n!)^2 2^{2n}3^{n}}$.
\end{questionparts}

Solution source

\begin{questionparts}
\item $\displaystyle (1-x)^{-1} = \sum_{n=0}^\infty x^n$,  $\displaystyle (1-x)^{-2} = \sum_{n=0}^\infty (n+1)x^n$,  $\displaystyle (1-x)^{-3} = \sum_{n=0}^\infty \frac{(n+2)(n+1)}{2}x^n$

\begin{align*}
&& \sum_{n=1}^{\infty} n2^{-n} &= \frac12\sum_{n=0}^{\infty}(n+1)2^{-n} \\
&&&= \frac12 (1-\tfrac12)^{-2} = 2 \\
\\
&& \sum_{n=1}^{\infty} nx^n&= x(1-x)^{-2} \\
\Rightarrow && \sum_{n=1}^{\infty} n^2x^{n-1}&= (1-x)^{-2}+2x(1-x)^{-3} \\
\Rightarrow && \sum_{n=1}^{\infty} n^22^{-n} &= \frac12 \left ( (1-\tfrac12)^{-2}+2\cdot \tfrac12 \cdot (1-\tfrac12)^{-3} \right) \\
&&&= \frac12 \left ( 4 +8\right) = 6
\end{align*}

\item By the generalised binomial theorem,
\begin{align*}
&& (1-x)^{-\frac12} &= 1 + \sum_{n=1}^{\infty} \frac{(-\tfrac12)\cdot(-\tfrac32)\cdots(-\tfrac12-n+1)}{n!}(-x)^n \\
&&&= 1 + \sum_{n=1}^{\infty} \frac{(-1)^n(\tfrac12)\cdot(\tfrac32)\cdots(\tfrac{2n-1}2)}{n!}(-x)^n \\
&&&= 1 + \sum_{n=1}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots (2n-1)}{2^nn!}x^n \\
&&&= 1 + \sum_{n=1}^{\infty} \frac{(2n)!}{2^nn! \cdot 2^n n!}x^n \\
&&&= 1 + \sum_{n=1}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\
&&&= \sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\
\end{align*}

\begin{align*}
&& \sum_{n=0}^\infty \frac{(2n)!} {(n!)^2 2^{2n}3^{n}} &= (1-\tfrac13)^{-\frac12} \\
&&&= \sqrt{\frac32} \\
\\
&& (1-x)^{-\frac12} &= \sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^n \\
\Rightarrow && \tfrac12(1-x)^{-\frac32} &= \sum_{n=0}^{\infty} \frac{n(2n)!}{2^{2n}(n!)^2}x^{n-1} \\
\Rightarrow && \sum_{n=1}^\infty \frac{n(2n)!} {(n!)^2 2^{2n}3^{n}} &= \frac16(1-\tfrac13)^{-3/2} \\
&&&= \frac16 \sqrt{\frac{27}{8}} = \frac14\sqrt{\frac{3}2}
\end{align*}

\end{questionparts}