10
$\begingroup$

How can we calculate the formula for the following series?

$\frac13+\frac29+\frac{1}{27}+\frac{2}{81}+\frac{1}{243}+\frac{2}{729}+\ldots$

I'm not quite sure what type of series it is anyway. It doesn't look geometric. Is there any strategy/procedure to follow in order to calculate the formula for a series of numbers?

$\endgroup$
2
  • 6
    $\begingroup$ Look at the sequence as made of two interwoven sequences, one going $1/3,1/27,1/243,...$ and the other going $2/9,2/81,2/729...$. Treat each separately using geometric series. $\endgroup$ – coffeemath Apr 4 '13 at 23:29
  • $\begingroup$ @coffeemath: same idea, two ways of looking at it :-) $\endgroup$ – robjohn♦ Apr 4 '13 at 23:38
14
$\begingroup$

If you are looking for the infinite sum, it is the sum of two geometric series: $$ \sum_{k=1}^\infty\frac1{3^k}+\sum_{k=1}^\infty\frac1{9^k}=\frac12+\frac18=\frac58 $$

$\endgroup$
10
$\begingroup$

If we add neighbouring pairs, we get $$\frac{5}{9}+\frac{5}{81}+\frac{5}{729}+\cdots.$$ This is an infinite geometric series with first term $\frac{5}{9}$ and common ratio $\frac{1}{9}$.

By the usual formula for the sum of an infinite geometric progression, the sum is $$\frac{5}{9}\frac{1}{1-\frac{1}{9}}.$$ This can be simplified to $\frac{5}{8}$.

$\endgroup$
6
$\begingroup$

$$a_n=\begin{cases}\frac{1}{3^n}&,\;\;\text{if $\,n\,$ is odd}\\{}\\\frac{2}{3^n}&,\;\;\text{if $\,n\,$ is even}\end{cases}\;=\ldots\;$$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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