Arithmetic - Geometric Series Duality

$\mathfrak{S}=\frac { 1 }{ 3 } +\frac { 2 }{ 9 } +\frac { 3 }{ 27 } +\frac { 4 }{ 81 } +\dots~~~~~~$ $\dfrac{1}{3}\mathfrak{S}=~~~~~~~~\frac { 1 }{9 } +\frac { 2 }{ 27} +\frac { 3 }{ 81 } +\frac { 4 }{ 243} +\dots$ Subtracting we get: $\dfrac{2}{3}\mathfrak{S}=\frac { 1 }{ 3 }+ \frac { 1 }{9 } +\frac { 1}{ 27} +\frac { 1 }{ 81 } +\frac { 1 }{ 243} +\dots$ $\color{#D61F06}{\mathcal{\text{(Infinite GP)}}}$ $\large \Rightarrow \mathfrak{S}=\dfrac{3}{2}(\dfrac{\frac{1}{3}}{1-\frac{1}{3}})=\dfrac{3}{4}$ $\huge\therefore~ 3\times 4=\boxed{\color{#007fff}{12}}$

$S=\frac { 1 }{ 3 } +\frac { 2 }{ 9 } +\frac { 3 }{ 27 } +\frac { 4 }{ 81 } +\dots \\ \\ \frac { S }{ 3 } =\frac { 1 }{ 9 } +\frac { 2 }{ 27 } +\frac { 3 }{ 81 } +\frac { 4 }{ 243 } +\dots \\ \\ S-\frac { S }{ 3 } =\frac { 1 }{ 3 } +\frac { 1 }{ 9 } +\frac { 1 }{ 27 } +\frac { 1 }{ 81 } +\dots \quad \left( \text{GP} \right)$

The GP can be simplified as

$\frac { 2 }{ 3 } S=\frac { \frac { 1 }{ 3 } }{ 1-\frac { 1 }{ 3 } } =\frac { 1 }{ 2 } \\ S=\frac { 3 }{ 4 }$

Therefore, the answer is $3 \cdot 4 = \boxed { 12 }$ .

This may be overcomplicating things... but I think it's a bit more rigorous than other solutions.

Start with the $m$ th partial sum $S_m = \sum_{n=0}^m \frac{n}{3^n} \\ S_m + \frac{m+1}{3^{m+1}} = \sum_{n=1}^{m+1} \frac{n}{3^n}\\ = \sum_{n=0}^{m} \frac{n+1}{3^{n+1}} = \frac{1}{3} \sum_{n=0}^{m} \frac{n+1}{3^n} = \frac{1}{3} \sum_{n=0}^{m} \Big( \frac{n}{3^n} + \frac{1}{3^n} \Big) = \frac{1}{3} S_m + \frac{1}{3} \sum_{n=0}^{m} \big(\frac{1}{3}\big)^n .\\$

So, just to rewrite the equation before doing some algebra:

$S_m + \frac{m+1}{3^{m+1}} = \frac{1}{3} S_m + \frac{1}{3} \sum_{n=0}^{m} \big(\frac{1}{3}\big)^n \\ 2 S_m = \sum_{n=0}^{m} \big(\frac{1}{3}\big)^n - \frac{m+1}{3^{m}}$

Now, letting $m \rightarrow \infty$ , we find that the first sum on the right is just an infinite GP and the second term vanishes:

$2 S = \frac{1}{1 - \frac{1}{3}} - 0 \\ S = \frac{3}{4}$

Starting with:

$\sum_{n=1}^{\infty} \frac{n}{3^{n}}$ = $\sum_{n=1}^{\infty} nx^{n}$ , where $x=$ $\frac{1}{3}$ , but:

$\sum_{n=1}^{\infty} nx^{n}$ = $x \times \sum_{n=1}^{\infty} nx^{n-1}$ = $x\times(\sum_{n=1}^{\infty} x^{n})'$ = $x\times(\frac{1}{1-x}-1)'$ = $\frac{x}{(1-x)^2}$

So as x= $\frac{1}{3}$ :

$\sum_{n=1}^{\infty} \frac{n}{3^{n}}$ = $\frac{\frac{1}{3}}{(1-\frac{1}{3})^2}$ = $\frac{3}{4}$ , so the answer is $3\times4=12$ .