Convergent, non-geometric, infinite sum as a fraction? How would you go about finding the sum of an infinite series that is known to be convergent, but is not geometric? 
The particular example I am interested in is as follows: $\sum_{j=0}^{\infty}\left [(6j+1)(\frac{1}{2})^{6j+2}\right ]$. 
The only catch to this is that I need to find the answer in fractional form. I've looked online and can't find anything useful relating to non-geometric series' or even to finding the fractional answer. 
 A: We have 
\begin{eqnarray*}
\sum_{j=0}^{\infty} x^j =\frac{1}{1-x}.
\end{eqnarray*}
Differentiate this
\begin{eqnarray*}
\sum_{j=0}^{\infty} j x^{j-1} =\frac{1}{(1-x)^2}.
\end{eqnarray*}
Your sum can be rewitten as
\begin{eqnarray*}
\frac{6}{256}\sum_{j=0}^{\infty} j \left( \frac{1}{64} \right)^{j-1} +\frac{1}{4}\sum_{j=0}^{\infty} \left( \frac{1}{64} \right)^j & = & \frac{3}{4} \frac{1}{(1-\frac{1}{64})^2} +\frac{1}{4} \frac{1}{(1-\frac{1}{64})} \\
&=& \frac{32}{1323} + \frac{16}{63} = \color{red}{\frac{368}{1323}}.
\end{eqnarray*}
EDIT : In more detail ...
\begin{eqnarray*}
S= \sum_{j=0}^{\infty} (6j+1) \left( \frac{1}{2} \right)^{6j+2} = 6 (\frac{1}{2})^2 \sum_{j=0}^{\infty}j \left( \left(\frac{1}{2} \right)^6 \right)^j +(\frac{1}{2})^2 \sum_{j=0}^{\infty} \left( \left(\frac{1}{2} \right)^6 \right)^j
\end{eqnarray*}
Now $(1/2)^6=1/64$ and the first sum needs the exponent to decrease by $1$ ... so
\begin{eqnarray*}
S= \frac{6}{4 \times 64}\sum_{j=0}^{\infty} j \left( \frac{1}{64} \right)^{j-1} +\frac{1}{4}\sum_{j=0}^{\infty} \left( \frac{1}{64} \right)^j & = & \frac{3}{2 \times 64} \frac{1}{(1-\frac{1}{64})^2} +\frac{1}{4} \frac{1}{(1-\frac{1}{64})} \\
& = & \frac{3}{2 \times 64} \left(\frac{64}{63}\right)^2 +\frac{1}{4} \frac{64}{63} \\
\end{eqnarray*}
