1. ## Infinite Series

Find the sum of:

$\displaystyle 1+\frac{1}{3}+\frac{2}{9}+\frac{1}{9}+\frac{4}{81} +\frac{5}{243}....$

I have tried to group the terms, to create two geometric series, but that does not work. I can see no pattern from one term to the next, although i have noted:

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

but I do not think this observation helps much. Any assistance is appreciated.

2. I would write it as $\displaystyle \displaystyle1 + \sum\limits_{k = 1}^\infty {\frac{k}{{3^k }}}$.

Recall that if $\displaystyle \displaystyle\sum\limits_{k = 1}^\infty {kx^k } = \frac{x}{{\left( {1 - x} \right)^2 }},\;\left| x \right| < 1$.

Let $\displaystyle x=\frac{1}{3}$

3. You are right!... the series can be written as...

$\displaystyle \displaystyle S= 1 + \sum_{n=1}^{\infty} n\ (\frac{1}{3})^{n}$ (1)

If we introduce the function...

$\displaystyle \displaystyle \sigma(x)= 1 + \sum_{n=1}^{\infty} n\ x^{n}= 1 + x\ \sum_{n=1}^{\infty} n\ x^{n-1} = 1+ x\ \frac{d}{dx} \sum_{n=0}^{\infty} x^{n}=$

$\displaystyle \displaystyle 1 + x\ \frac{d}{dx} \frac{1}{1-x} = 1+\frac{x}{(1-x)^{2}}$ (2)

... is easy to derive that is...

$\displaystyle \displaystyle S=\sigma(\frac{1}{3}) = \frac{7}{4}$ (3)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

4. Hello, rtblue!

Here is an elementary solution . . .

Find the sum of: .$\displaystyle \displaystyle S \;=\;1+\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\fr ac{4}{3^4}+\frac{5}{3^5}\ + \hdots$

$\displaystyle \begin{array}{ccccccc} \text{We have:} & S &=& 1 + \frac{1}{3} + \frac{2}{3^2} + \frac{3}{3^3} + \frac{4}{3^4} + \hdots & [1] \\ \text{Multiply by}\frac{1}{3}\!: & \frac{1}{3}S &=& \quad\;\;\;\frac{1}{3} + \frac{1}{3^2} + \frac{2}{3^3} + \frac{3}{3^4} + \hdots & [2] \\\end{array}$

Subtract [1] - [2]: .$\displaystyle \frac{2}{3}S \;=\;1 + \frac{1}{3^2} + \frac{1}{3^3} + \frac{1}{3^4} + \frac{1}{3^5} + \hdots$

. . . . . . . . . . . . . $\displaystyle \frac{2}{3}S \;=\;1 +\frac{1}{3^2}\underbrace{\left(1 + \tfrac{1}{3} + \tfrac{1}{3^2} + \tfrac{1}{3^3} + \hdots \right)}_{\text{Infinite series}}\;\;[3]$

The infinite series has the sum: .$\displaystyle \dfrac{1}{1-\frac{1}{3}} \:=\:\dfrac{1}{\frac{2}{3}} \:=\:\frac{3}{2}$

Substitute into [3]: .$\displaystyle \frac{2}{3}S \;=\;1 + \frac{1}{9}\left(\frac{3}{2}\right) \;=\;1 + \frac{1}{6} \;=\;\frac{7}{6}$

. . . . . . Therefore: . .$\displaystyle S \;=\;\frac{7}{4}$