Welcome to our community

Be a part of something great, join today!

[SOLVED] sum 1/3

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
2\frac{1}{3} + \frac{1}{3^2} + 2\frac{1}{3^3} + \frac{1}{3^4} + 2\frac{1}{3^5} + \cdots = \frac{1}{3}\sum_{n = 0}^{\infty}\left(\frac{1}{3}\right)^n
$$
I am stuck on what to add in to account for the 2 at every other term.
 

Fantini

"Read Euler, read Euler." - Laplace
MHB Math Helper
Feb 29, 2012
342
Re: infinite series summation problem

What if

$$2 \frac{1}{3} + \frac{1}{3^2} + 2 \frac{1}{3^3} + \frac{1}{3^4} + 2 \frac{1}{3^5} + \cdots = \left( \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \frac{1}{3^5} + \cdots \right) + \left( \frac{1}{3} + \frac{1}{3^3} + \frac{1}{3^5} + \cdots \right)$$

$$= \frac{1}{3} \left( 1 + \frac{1}{3} + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \cdots \right) + \frac{1}{3}\left( 1 + \frac{1}{3^2} + \frac{1}{3^4} + \cdots \right).$$

The first parenthesis is a geometric series with ratio $\frac{1}{3}$ and the second with ratio $\frac{1}{9}$.

This doesn't seem right though, because the sum would be $\frac{7}{8}$, whereas on the right side you'd get $\frac{1}{2}$. Perhaps this change of terms in the series I did is wrong, but I doubt that because the series on the right, which it is supposed to be, is absolutely convergent, being a geometric series with ratio $\frac{1}{3}$.

I hope this helps spring the discussion into the right direction. :D
 
Last edited by a moderator:

MarkFL

Administrator
Staff member
Feb 24, 2012
13,775
I think I would rewrite the sum as:

$\displaystyle \frac{2}{3}\sum_{k=0}^{\infty}\left(\frac{1}{9} \right)^k+\frac{1}{9} \sum_{k=0}^{\infty}\left(\frac{1}{9} \right)^k=\frac{7}{9}\sum_{k=0}^{\infty}\left( \frac{1}{9} \right)^k$