0
$\begingroup$

This question already has an answer here:

I'm trying to take the limit of a series involving $\sum_{j=1}^\infty\frac{1}{3^j}$ and thinking that this might have a partial sum representation.

Here it says that $\sum_{i=1}^n \frac{1}{3^{i-1}} = \frac{3}{2}(1-\frac{1}{3^n})$ is the partial sum representation for $\sum_{i=1}^n \frac{1}{3^{i-1}}$:
http://tutorial.math.lamar.edu/Classes/CalcII/ConvergenceOfSeries.aspx

But how is this derived?

$\endgroup$

marked as duplicate by user147263, user1551, Strants, N. F. Taussig, jameselmore Sep 20 '15 at 17:58

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ There is no "partial fraction" here. $\endgroup$ – Did Sep 14 '15 at 7:55
  • $\begingroup$ @Did Ah correct. It's "partial sum". $\endgroup$ – mavavilj Sep 14 '15 at 7:56
  • 1
    $\begingroup$ By the by, did you try the obvious before asking this? en.wikipedia.org/wiki/Geometric_series $\endgroup$ – Did Sep 14 '15 at 7:56
  • $\begingroup$ @Did No, because I was mistakenly looking for "partial fractions", since the series I'm dealing with has actually multiplicative terms, where $\sum_{j=1}^\infty\frac{1}{3^j}$ is one such term. $\endgroup$ – mavavilj Sep 14 '15 at 7:58
  • $\begingroup$ Or of Partial sum of a geometric series $\endgroup$ – user147263 Sep 14 '15 at 11:55
3
$\begingroup$

Are you familiar with the formula for the sum of the first n terms of a geometric series $$\sum\limits_{k=0}^{n-1} r^k=\frac{1-r^n}{1-r}$$ or something similar?

Using this formula you easily get: $$\sum\limits_{i=1}^{n} \frac{1}{3^{i-1}} =\sum\limits_{i=1-1}^{n-1} \frac{1}{3^{(i+1)-1}} =\sum\limits_{i=0}^{n-1} \frac{1}{3^i} =\sum\limits_{i=0}^{n-1} \left(\frac{1}{3}\right)^{i}$$

Can you take it from here?

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.