$\displaystyle (\frac{1}{3^3}+\frac{1}{3^4})+(\frac{2}{3^5}+\frac {2}{3^6})+(\frac{3}{3^7}+\frac{3}{3^8})+(\frac{4}{ 3^9}+\frac{4}{3^{10}})+.....$
Put:
$\displaystyle S(x)=\sum_{n=0}^{\infty} x^n=\frac{1}{1-x}$
Now differentiate term by term:
$\displaystyle S'(x)=\sum_{n=1}^{\infty} n x^{n-1}=\frac{1}{(1-x)^2}$
so:
$\displaystyle xS'(x)=\sum_{n=1}^{\infty} n x^n=\frac{x}{(1-x)^2}$
Now put $\displaystyle x=1/9$.
(also see attachment)
RonL
Without differentiating
We want to find $\displaystyle G(x)=\sum_{k=0}^{\infty}{\left(k\cdot{x^k}\right)}$ ( By the root test you can see that the series converges for $\displaystyle |x|<1$)
Now we consider:
$\displaystyle x\cdot{G(x)}=\sum_{k=0}^{\infty}{\left(k\cdot{x^{k +1}}\right)}=\sum_{k=1}^{\infty}{(k-1)\cdot{x^k}}$
So: $\displaystyle G(x)-x\cdot{G(x)}=\sum_{k=0}^{\infty}{\left(k\cdot{x^k} \right)}-\sum_{k=1}^{\infty}{(k-1)\cdot{x^k}}=\sum_{k=1}^{\infty}{(k-k+1)\cdot{x^k}}=\sum_{k=1}^{\infty}{x^k}=\frac{x}{ 1-x}$ whenever $\displaystyle |x|<1$ (Because it is a geometric series)
Thus: $\displaystyle (1-x)\cdot{G(x)}=\frac{x}{1-x}$ and we obtain:$\displaystyle G(x)=\frac{x}{(1-x)^2}$
Hello, perash!
$\displaystyle S \;=\;\left(\frac{1}{3^3}+\frac{1}{3^4}\right)+\lef t(\frac{2}{3^5}+\frac{2}{3^6}\right)+\left(\frac{3 }{3^7}+\frac{3}{3^8}\right)+\left(\frac{4}{3^9}+\f rac{4}{3^{10}}\right)+ \cdots$
We have: . . . . . . . . . .$\displaystyle S \;=\;\frac{4}{3^4} + \frac{8}{3^6} + \frac{12}{3^8} + \frac{16}{3^{10}} + \cdots$
Factor: . . . . . . . . . . . $\displaystyle S\;=\;\frac{4}{3^4}\left[1 + \frac{2}{3^2} + \frac{3}{3^4} + \frac{4}{3^4} + \cdots\right]\;\;{\color{blue}[1]}$
Multiply by $\displaystyle \frac{1}{3^2}:$ . . . . . $\displaystyle \frac{1}{9}S \;=\;\frac{4}{3^4}\left[\qquad \frac{1}{3^2} + \frac{2}{3^4} + \frac{3}{3^6} + \cdots\right]\;\;{\color{blue}[2]} $
Subtract [2] from [1]: .$\displaystyle \frac{8}{9}S \;=\;\frac{4}{3^4}\underbrace{\left[1 + \frac{1}{3^2} + \frac{1}{3^4} + \frac{1}{3^6} + \cdots\right]}_{\text{geometric series: }a = 1,\:r = 1/9} $
Hence, we have: . . . . $\displaystyle \frac{8}{9}S \;=\;\frac{4}{81}\left[\frac{9}{8}\right] \quad\Rightarrow\quad\boxed{S \:=\:\frac{1}{16}}$