Problem source

Sum the following infinite series.
\begin{questionparts}
	\item \[ 1 + 
\frac13 \bigg({\frac12}\bigg)^2 +\frac15\bigg(\frac12\bigg)^4
+ \cdots + \frac{1}{2n+1} \bigg(\frac12\bigg)^{2n} + \cdots \] .
\item \[
2 -x -x^3 +2x^4 - \cdots + 2x^{4k} - x^{4k+1} - x^{4k+3} +\cdots \]
where $|x| < 1$.
\item  \[ \sum _{r=2}^\infty
 {r\,  2^{r-2} \over 3^{r-1} } \].
\item \[ \sum_{r=2}^\infty
 {2 \over r(r^2-1) } \].
 \end{questionparts}

Solution source

\begin{questionparts}
\item \begin{align*}
&& \sum_{i=0}^{\infty} x^{2i+1}&= \frac{x}{1-x^2}  \\
\Rightarrow &&&=\frac12 \left ( \frac{1}{1-x} - \frac{1}{1+x} \right) \\
\underbrace{\Rightarrow}_{\int} && \sum_{i=0}^{\infty} \frac{1}{2i+1} x^{2i+2} &= \frac12 \left ( -\ln (1-x) - \ln(1+x) \right) \\
\underbrace{\Rightarrow}_{x = 1/2} && \sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i+2} &= -\frac12 \ln \frac12 - \frac12 \ln \frac32 \\
&&&= -\frac12 \ln \frac34 \\
&&\frac14\sum_{i=0}^\infty \frac{1}{2i+1} \left (\frac12 \right)^{2i}&= \frac12 \ln \frac43 \\
\Rightarrow&& S &= 2 \ln \frac43
\end{align*}
\item \begin{align*}
\sum_{k=0}^{\infty} \left (2x^{4k} - x^{4k+1} - x^{4k+3} \right) &= \sum_{k=0}^{\infty} \left (2 - x^{1} - x^{3} \right) x^{4k} \\
&= \frac{2-x-x^3}{1-x^4} \\
&= \frac{(1-x)(2+x+x^2)}{(1-x)(1+x+x^2+x^3)} \\
&= \frac{2+x+x^2}{1+x+x^2+x^3}
\end{align*}
\item \begin{align*}
&& \frac{1}{(1-x)^2} &= \sum_{r=0}^{\infty} r x^{r-1} \\
\Rightarrow && 9 &= \sum_{r=1}^{\infty} r \left ( \frac23 \right)^{r-1} \\
\Rightarrow &&  \sum_{r=2}^{\infty} r \left ( \frac{2^{r-2}}{3^{r-1}} \right) &= \frac12 \left ( 9 - 1 \right) \\
&&&= 4
\end{align*}

\item \begin{align*}
&& \frac{2}{r(r^2-1)} &= \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \\
\Rightarrow && \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{2}{r} + \frac{1}{r+1} \right) &= \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} - \frac{1}{r} + \frac{1}{r+1} \right) \\
&&&=  \sum_{r=2}^{\infty} \left ( \frac{1}{r-1} - \frac{1}{r} \right)-\sum_{r=2}^{\infty} \left ( \frac{1}{r} - \frac{1}{r+1} \right) \\
&&&= 1 - \frac12 \\
&&&= \frac12
\end{align*}

\end{questionparts}