Use the Cauchy product to find the sum of $\sum_{n=1}^\infty \frac{n}{3^n}$ I proved that the serie
$\displaystyle{\sum_{n = 1}^{\infty}{n \over 3^{n}}}$ is convergent, but I want to find the value of the sum using the Cauchy product. Any suggestion ?.
 A: Let $$a_n=\frac 1{3^{n}} \qquad b_n=\frac 1{3^{n}}$$
then
$$\left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)\times \left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)=\sum_k\sum_{l=0}^k \frac 1{3^{l}}\times \frac{3^l}{3^k}$$
$$\left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)^2=\sum_k\frac 1{3^k}\times(k+1)$$
$$ \left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)^2=x+\left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)$$
Where $$x=\sum_k \frac{k}{3^k}$$
Geometric series (can't we use this particular case at least?): $$\left(\sum_{i=0}^\infty \frac 1{3^{n}}\right)=\frac 32$$
Hence $$\frac 94=\frac 32 +x$$
Finally $$x=\frac 34$$
A: You don't need the Cauchy product. It is sufficient to note that $$\sum_{n\geq0}x^{n}=\frac{1}{1-x},\,\left|x\right|<1
 $$ hence taking the derivative $$\sum_{n\geq0}nx^{n-1}=\frac{1}{\left(1-x\right)^{2}}
 $$ so $$\sum_{n\geq0}nx^{n}=\sum_{n\geq1}nx^{n}=\frac{x}{\left(1-x\right)^{2}}
 $$ hence taking $x=\frac{1}{3}
 $ we have 
$$\sum_{n\geq1}\frac{n}{3^{n}}=\color{red}{\frac{3}{4}}.$$
A: Use $\sum_{n\geq 0}x^{n}=\frac{1}{1-x}$ for $|x|<1$. 
\begin{align}
\frac{1}{(1-x)^{2}}=\left(\sum_{n\geq 0} x^{n}\right)^{2}=\sum_{m,n\geq0}x^{m+n}=\sum_{r\geq0}(r+1)x^{r}
\end{align}
(Here we use Cauchy product.) Then we have 
\begin{align}
\sum_{r\geq0}rx^{r}=\sum_{r\geq0}(r+1)x^{r}-\sum_{r\geq0}x^{r}=\frac{1}{(1-x)^{2}}-\frac{1}{1-x}
\end{align}
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\color{#f00}{\sum_{n = 1}^{\infty}{n \over 3^{n}}} & =
\sum_{n = 1}^{\infty}{n \choose n - 1}\pars{1 \over 3}^{n} =
{1 \over 3}\sum_{n = 0}^{\infty}{n + 1 \choose n}\pars{1 \over 3}^{n}
\\[4mm] & =
{1 \over 3}\sum_{n = 0}^{\infty}
{-n - 1 -n - 1 \choose n}\pars{-1}^{n}\pars{1 \over 3}^{n} =
{1 \over 3}\sum_{n = 0}^{\infty}{-2 \choose n}\pars{-\,{1 \over 3}}^{n}
\\[4mm] & =
{1 \over 3}\bracks{1 + \pars{-\,{1 \over 3}}}^{-2} =
\color{#f00}{3 \over 4}
\end{align}
A: Another approach that uses the geometric series is to recognize that
$$\begin{align}
\sum_{n=1}^N \frac{n}{3^n}&=\sum_{n=1}^N \frac{1}{3^n}\sum_{m=1}^n(1)
\end{align}$$
Now, interchanging the order of summation reveals
$$\begin{align}
\sum_{n=1}^N \frac{n}{3^n}&=\sum_{m=1}^N \sum_{n=m}^N\frac{1}{3^n}\\\\
&=\sum_{m=1}^N \frac{(1/3)^m-(1/3)^{N+1}}{1-1/3}\\\\
&=\frac32 \left(\frac{(1/3)-(1/3)^{N+1}}{1-1/3}-N(1/3)^{N+1}\right)\\\\
&\to \frac34\,\,\text{as}\,\,N\to \infty
\end{align}$$
