Thread: Infinite Series

Evaluate:

$\displaystyle \frac{1}{2}+\frac{1}{4}+\frac{1}{4}+\frac{3}{16}+\ frac{5}{32}...$

I know the formula for an infinite series:

$\displaystyle \frac{a}{1-r}$


But i cannot seem to find the rate. I have tried grouping the terms, and doing two infinite series, but I still cannot find a pattern. Any help is appreciated.

Hello, rtblue!

Is there a typo?
Shouldn't the third fraction be $\displaystyle \frac{1}{8}\,?$

$\displaystyle \displaystyle \text{Evaluate: }\:S \:=\:\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{3}{ 16}+\frac{5}{32} + \hdots$

We have: .$\displaystyle \displaystyle S \;=\;\left(\frac{1}{2} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{3}{16} + \frac{5}{32} + \frac{7}{64} + \hdots\right)$

. . . . ,. . . $\displaystyle \displaystyle S \;=\;\frac{3}{8} + \frac{1}{8}\underbrace{\left(1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \hdots\right)}_{\text{This is }T}$ .[1]


$\displaystyle \begin{array}{ccccc}
\text{We have:} & T &=& 1 + \frac{3}{2} + \frac{5}{4} + \frac{7}{8} + \frac{9}{16} + \hdots & [2]\\ \\[-3mm]
\text{Multiply by }\frac{1}{2}\!: & \frac{1}{2}T &=& \quad\;\; \frac{1}{2} + \frac{3}{4} + \frac{5}{8} + \frac{7}{16} + \hdots & [3]\end{array}$

Subtract [2]-[3]: .$\displaystyle \frac{1}{2}T \;=\;\;1 + \frac{2}{2} + \frac{2}{4} + \frac{2}{8} + \frac{2}{16} + \hdots$

. . . . . . . . . . . . .$\displaystyle \frac{1}{2}T \;=\; 1 + \underbrace{\left(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \hdots\right)}_{\text{This sums to 2}}$

. . . . . . . . . . . . .$\displaystyle \frac{1}{2}T \;=\;1 + 2 \quad\Rightarrow\quad T \:=\:6$


Substitute into [1]: .$\displaystyle \displaystyle S \;=\;\frac{3}{8} + \frac{1}{8}(6) \;=\;\frac{9}{8}$

I do not believe there is a typo, but I am taking the question from an online source, so there might be some error there. Anyways, the answer given to my question is 2. Also, i have noticed that the numerators form the fibonacci series. I am not sure if this helps though.

Actually i have just found a formula, for a problem like this, involving the fibonacci series:

$\displaystyle \frac{r}{r^2-(r+1)}$

for:

$\displaystyle \frac{1}{r}+\frac{1}{r^2}+\frac{2}{r^3}+\frac{3}{r ^4}...$

and the numerators form the Fibonacci series.

note that the second 1/4 can be expressed as 2/8.

after substituting r, we get the sum to be 2.