Good morning, i have a problem when i go to calculate the partial sum of this series:

$S = 2+\frac{2}{3}+\frac{2}{9}+\frac{2}{27}+...+\frac{2}{3^{n-1}}$

I make this:

If this an geometric series then $a=2$ and $r=\frac{1}{3}$


$S={\displaystyle \sum_{i=1}^{n}2(\frac{1}{3}})^{i-1}$

but, i cannot calculate the partial sum, please help me!

  • 1
    $\begingroup$ I´ve made an edit, because series is plural as well as singular. $\endgroup$ – callculus May 8 '16 at 15:19
  • $\begingroup$ not problem man, thanks @callculus $\endgroup$ – Bvss12 May 8 '16 at 15:25

Geometric series: $$a+ar+ar^2+\cdots+ar^{n-1}=\frac{a(r^n-1)}{r-1}$$

Now can you apply that to your series? (Hint: $a$ is the first term, $r$ is the common ratio.)

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  • $\begingroup$ oh, interesting... then my partial sum is: $-3\, \left( 1/3 \right) ^{n-1}+3$, or well, this: $\frac{2((\frac{1}{3})^{n-1}-1)}{\frac{1}{3}-1}$ @Kenny Lau $\endgroup$ – Bvss12 May 8 '16 at 15:09
  • $\begingroup$ I think it should be $-3(1/3)^n+3$ instead... $\endgroup$ – Kenny Lau May 8 '16 at 15:11
  • $\begingroup$ But anyways, you get it. $\endgroup$ – Kenny Lau May 8 '16 at 15:11
  • $\begingroup$ are you sure, i say n-1 because the last term... $\endgroup$ – Bvss12 May 8 '16 at 15:12

Hint : $S={\displaystyle \sum_{i=1}^{n}2(\frac{1}{3}})^{i-1}={\displaystyle 6\sum_{i=1}^{n}(\frac{1}{3}})^{i}$, and you recognize a geometric series.

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  • $\begingroup$ Upvote for the hint. But series is plural as well as singular. $\endgroup$ – callculus May 8 '16 at 15:18
  • $\begingroup$ @callculus Oh I always had a doubt, so I should "you recognize geometric series" right ? $\endgroup$ – Bérénice May 8 '16 at 15:19
  • $\begingroup$ I´m really not an expert in English. But I would say:"...and you recognize $a$ geometric series". $\endgroup$ – callculus May 8 '16 at 15:24
  • $\begingroup$ oh ok well I will pay attention in the other topics to be sure how it works. Thank you $\endgroup$ – Bérénice May 8 '16 at 15:27

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