# Partial sums for $\sum_{j=1}^\infty\frac{1}{3^j}$ and similar. [duplicate]

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?

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

• There is no "partial fraction" here. – Did Sep 14 '15 at 7:55
• @Did Ah correct. It's "partial sum". – mavavilj Sep 14 '15 at 7:56
• By the by, did you try the obvious before asking this? en.wikipedia.org/wiki/Geometric_series – Did Sep 14 '15 at 7:56
• @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. – mavavilj Sep 14 '15 at 7:58
• – user147263 Sep 14 '15 at 11:55

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: $$\sum\limits_{i=1}^{n} \frac{1}{3^{i-1}} =\sum\limits_{i=1-1}^{n-1} \frac{1}{3^{(i+1)-1}} =\sum\limits_{i=0}^{n-1} \frac{1}{3^i} =\sum\limits_{i=0}^{n-1} \left(\frac{1}{3}\right)^{i}$$