Problem source

Obtain the sum to infinity of each of the following series. 
\begin{questionparts}
\item $1{\displaystyle +\frac{2}{2}+\frac{3}{2^{2}}+\frac{4}{2^{3}}+\cdots+\frac{r}{2^{r-1}}+\cdots;}$
\item $1{\displaystyle +\frac{1}{2}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{2^{2}}+\cdots+\frac{1}{r}\times\frac{1}{2^{r-1}}+\cdots;}$
\item ${\displaystyle \dfrac{1\times3}{2!}\times\frac{1}{3}+\frac{1\times3\times5}{3!}\frac{1}{3^{2}}+\cdots+\frac{1\times3\times\cdots\times(2k-1)}{k!}\times\frac{1}{3^{k-1}}+\cdots.}$
\end{questionparts}
[Questions of convergence need not be considered.]

Solution source

\begin{questionparts}
\item \begin{align*}
&& \frac1{1-x} &= \sum_{r=0}^{\infty} x^r \\
\underbrace{\Rightarrow}_{\frac{\d}{\d x}} && \frac{1}{(1-x)^2} &= \sum_{r=0}^\infty rx^{r-1} \\
\underbrace{\Rightarrow}_{x = \frac12} && 4 &= \sum_{r=0}^{\infty} \frac{r}{2^{r-1}}
\end{align*}

\item \begin{align*}
&& \frac1{1-x} &= \sum_{r=1}^{\infty} x^{r-1} \\
\underbrace{\Rightarrow}_{\int} && -\ln (1-x) &= \sum_{r=1}^{\infty} \frac1{r} x^r \\
\underbrace{\Rightarrow}_{x = \frac12} && \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r}} \\
\Rightarrow && 2 \ln 2 &= \sum_{r=1}^{\infty} \frac1{r } \times \frac{1}{ 2^{r-1}} \\
\end{align*}

\item \begin{align*}
&& (1-x)^{-1/2} &= 1 + \frac{(-\tfrac12)}{1!} (-x) +\frac{(-\tfrac12)(-\tfrac32)}{2!}(-x)^2 + \cdots \\
&&&= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{2^rr!} x^r \\
\underbrace{\Rightarrow}_{x = \frac23} && \sqrt{3} &= \sum_{r=0}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^r} \\
&&&= 1 + \frac{1}{1!} \frac23 +  \frac13 \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\
\Rightarrow && 3\sqrt{3}-5 &= \sum_{r=2}^{\infty} \frac{1 \cdot 3 \cdot 5 \cdots \cdot (2r-1)}{r!} \frac1{3^{r-1}} \\
\end{align*}
\end{questionparts}