$$\frac{1}{4}, \frac{1}{12}, \frac{1}{36}, \frac{1}{108}, \ldots,\frac{1}{2916}$$

I am trying to calculate the sum of this geometric series. Here's what I've got so far.

$a = \frac{1}{4}, r = \frac{1}{3}$ and $n = 7$

So $\sum_{n=7}\frac{1}{4}(\frac{1}{3})^{n}$

Somehow equals $\frac{1093}{2916}$ according to my book?

Here are my questions:

  1. How do I get the sum of this geometric series
  2. Does my work look correct?
  3. How can I find a way to calculate the number of terms in the series that isn't "brute forcing"? This is quite inelegant.


  • 1
    $\begingroup$ You have a lot of questions in this place that show you how to sum a geometric series. See the right panel, the related questions to this one. $\endgroup$ – Masacroso Mar 27 '16 at 4:05
  • $\begingroup$ I'm having a hard time finding the answer - plenty of abstract stuff out there but not any lists. Even my textbook doesn't show the work. $\endgroup$ – Don Shrinkle Mar 27 '16 at 4:08

The partial sum of a geometric series is

$$\sum_{k=0}^{n-1}ar^k = a\frac{1-r^n}{1-r}$$

so take $a=1/4, r=1/3, n=7$ to get

$$S = \left(\frac{1}{4}\right)\frac{1-(1/3)^7}{1-(1/3)} = \frac{1093}{2916}.$$

  • $\begingroup$ You sir, are great! $\endgroup$ – Don Shrinkle Mar 27 '16 at 4:18

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