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.
$\endgroup$ 24 Answers
$\begingroup$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?
$\endgroup$ 3 $\begingroup$Hint:
Think of the high school identity: $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dotsm+ab^{n-2}+b^{n-1}).$$
$\endgroup$ $\begingroup$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!
$\endgroup$ 1 $\begingroup$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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