How do you find the equation from the series expansion

Question

How do you find the equation which is a Taylor series expansion of $$\displaystyle 1+\frac{2}{3^2}+\frac{3}{3^3} + \ldots + \frac{k}{3^k} \ldots$$

I believe the Taylor series expansion : $\displaystyle \frac13+\frac{2}{3^2}+\frac{3}{3^3} + \ldots + \frac{k}{3^k} \ldots$ is the expansion of $\displaystyle \frac{x}{(1-x)^2}$.

Where does the $1$ come from then? How do I find what the original equation is?

Answer 1

Let $\displaystyle s(x)=\sum_{k=1}^\infty 3^{-xk}=\dfrac{1}{3^{x}-1}$ where $x>0$, then find $s'(x)$ and set $x=1$.

Answer 2

To find the series expansion of $x/(1-x)^2$, you could write it as $x[1/(1-x)][1/(1-x)]$ and use the geometric series expansion to get $$x(1+x+x^2+x^3+\cdots)(1+x+x^2+x^3+\cdots).$$ Now identify the $x^n$-coefficient in this product.