Arithmetic over Geometric

Let $S = \sum_{1}^{\infty} \frac{n}{3^{n}} = \frac{1}{3} + \frac{2}{9} + \frac{3}{27} + \frac{4}{81} + ......$ .

Then $\frac{1}{3} S = \frac{1}{9} + \frac{2}{27} + \frac{3}{81} + ......$ .

Thus $S - \frac{1}{3}S = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + ....$

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

So the desired sum is $2S = \frac{3}{2}$ , and thus $a - b = 3 - 2 = \boxed{1}$ .

By expanding out the sum and letting it equal $S$ , we have: $S = \frac{2}{3} + \frac{4}{9} + \frac{6}{27} + \frac{8}{81}\ldots$ By multiplying both sides by $\frac{3}{2}$ , the equation more simply becomes $\frac{3S}{2} = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27}\ldots$

Now consider the expansion $(1-x)^{-2}= 1 + 2x + 3x^2 + 4x^3 \ldots$ , by setting $x = \frac{1}{3}$ , It simply expands to $(1-\frac{1}{3})^{-2} = 1 + \frac{2}{3} + \frac{3}{9} + \frac{4}{27}\ldots = \frac{2S}{3}$ Hence, $(\frac{2}{3})^{-2} = \frac{3S}{2} \implies S = \frac{3}{2}$

There is a very simple way to solve this. That is using euclidean algorithm. The Euclidean algorithm is based on the principle that the greatest common divisor of two numbers does not change if the larger number is replaced by its difference with the smaller number.

So in this case, a-b= 1 because and b are co-prime integers (GCD=1)