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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.

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  • $\begingroup$ You surely know $\sum x^j=1/(1-x)$ for $|x|<1$, If you also knew $\sum jx^j$ you'd be in business! $\endgroup$ – Lord Shark the Unknown Jun 23 '17 at 18:37
  • $\begingroup$ Well then, it seems I just didn't know what to search for! I just found that $\sum(j-1)x^j = \frac{x^2}{(x-1)^2}$ Thank you! $\endgroup$ – Strafe Ae Jun 23 '17 at 18:43
  • $\begingroup$ Hmm... Just realized that that may not work when $j$ is a linear term as in the case above where it is $6j+2$. $\endgroup$ – Strafe Ae Jun 23 '17 at 18:49
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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*}

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  • $\begingroup$ Can you explain why we can rewrite that way? $\endgroup$ – Strafe Ae Jun 23 '17 at 19:05
  • $\begingroup$ Sure ... give a few minutes to edit in some more details $\endgroup$ – Donald Splutterwit Jun 23 '17 at 19:07
  • $\begingroup$ It took me a while, but I think I was able to follow it. I did notice a few typos though. One you already fixed, but the other is where you first write it as two sums; the first sum should be multiplied by $j$. $\endgroup$ – Strafe Ae Jun 23 '17 at 20:02

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