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I need to find a formula for the nth partial sum of the series:

$ 2 + \frac 23 + \frac 29 + \frac {2}{27} + .... + \frac {2}{3^{n-1}} + ... $

then I need to use the formula to find the series' sum if the series converges.

The answer from the back of the book is (the formula):

$$ \frac {2(1- (\frac 13) ^ n)}{1 - \frac 13} $$

And the series' sum is 3. Any help would be appreciated. Struggling to figure out how to get the formula.

--Edited because I messed up MathJax, fixed now.

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  • $\begingroup$ You should have a look to geometric progressions. en.m.wikipedia.org/wiki/Geometric_progression. $\endgroup$
    – mathcounterexamples.net
    Nov 2, 2015 at 19:16
  • $\begingroup$ Let $S=S(n)$ be the sum up to $\frac{2}{3^{n-1}}$. Then $\frac{1}{3}S$ is equal to the sum $\frac{2}{3}+\frac{2}{9}+\cdots +\frac{2}{3^n}$. Subtract from $S$, noting the beautiful cancellation. We get $S-\frac{1}{3}S=2-\frac{2}{3^n}$. Now can you finish? $\endgroup$
    – André Nicolas
    Nov 2, 2015 at 19:18

4 Answers 4

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Your series is a G.P. series with first term $2$ and common ratio $\frac{1}{3}$. So according to the G.P. sum formula, $S_n=\frac{2\left(1-(\frac{1}{3})^n\right)}{1-\frac{1}{3}}=3\left(1-(\frac{1}{3})^n\right)$. As $n\to\infty$, we can easily check that $S_n\to 3$.

Now do you know the G.P. formula?

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  • $\begingroup$ The G.P. formula is $ s_n = \frac {a(1-r^n)}{1-r} $ I believe? $\endgroup$
    – bankey
    Nov 2, 2015 at 19:18
  • $\begingroup$ Ahh, I understand. I didn't think to use that formula even though it was staring right at me. And then the ratio would be found by taking $ s_{n+1} / s_n $ . Thanks all! $\endgroup$
    – bankey
    Nov 2, 2015 at 19:22
  • $\begingroup$ Yes you're right. $\endgroup$
    – SchrodingersCat
    Nov 2, 2015 at 19:24
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Hint:

Think of the high school identity: $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dotsm+ab^{n-2}+b^{n-1}).$$

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First, it may be helpful to factor out $2$ from your series; it will only clutter up your actual work. So really, you are out for a formula of the $n^{th}$ partial sum of $\sum_{n=0}^\infty \frac{1}{3^n}$. As the comment on your post points out, this is just a special instance of a geometric progression $\sum_{n=0}^\infty x^n$ where $|x|<1$. Now, you could just appeal to the generally known formulas for this sort of thing, but I find it helpful to not have to memorize too many things; to this end, what we really want to do is convert this: $$ 1+x+x^2+\cdots+x^n $$ into something without dots. The best way to try to do this is multiply by another polynomial, and shoot for lots of cancellation. If you recall things like the difference of cubes formula from high school, the guess is immediate. $$ (1-x)\cdot(1+x+x^2+\cdots+x^n)=1+x-x+x^2-x^2+\cdots+x^n-x^n-x^{n+1}=1-x^{n-1} $$ So, you see that after division by $1-x$, your partial sum looks like $\frac{1-x^{n+1}}{1-x}$. Now plug in your specific value $x=1/3$ and multiply by $2$, and you are done!

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  • $\begingroup$ I would just like to add that while this method may seem roundabout, it can help you avoid making silly mistakes; I've seen many students blindly apply the GP formula and get questions wrong because they didn't verify that the summation index started from 0! $\endgroup$
    – Tenno
    Nov 2, 2015 at 19:27
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If the $n$th term in a series is $\lambda a^n$ then $$s_n = \sum_{i=0}^n \lambda a^i$$ $$a s_n = \sum_{i=1}^{n+1} \lambda a^{i}$$ $$s_n - a s_n = (1-a)s_n = \lambda - \lambda a^{n+1}$$ So that $$\sum_{i=0}^n \lambda a^i = \lambda\frac{1- a^{n+1}}{1-a}$$

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