Evaluate the sum \[\frac{1}{3^1} + \frac{2}{3^2} + \frac{3}{3^3} + \cdots + \frac{k}{3^k} + \cdots \]

I know it sums to 0.75    but I do not know how to compute that result ....  anyone?

Still working on a rigorous approach, but I used 2 identities that led to $\frac{1}{(1-x)^2} - \frac{1}{1-x} = (1x^0 + 2x^1 + 3x^2 + 4x^3 + \cdots) - (1x^0 + x^1 + x^2 + x^3 + \cdots) = x^1 + 2x^2 + 3x^3 + \cdots.$ Maybe this will help?

Got nowhere, but I think you should do it like this: https://ibb.co/ZJf62Cg

Approach 1: Manipulation

Let sum = S.

$$S=\frac{1}{3^1}+\frac{1}{3^2}+\frac{2}{3^2}+\frac{1}{3^3}+\frac{3}{3^3}+...=\left(\frac{1}{3^1}+\frac{1}{3^2}+...\right)+\frac{S}{3}=\frac{1}{2}+\frac{S}{3}$$

Hence $S=\frac{S}{3}+\frac{1}{2}\implies \frac{2S}{3}=\frac{1}{2}\implies \boxed{S=\frac{3}{4}}$.

Approach 2: Calculus

Take the series $\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n$. Differentiating yields $\frac{1}{(1-x)^2}=\sum_{n=0}^{\infty} nx^{n-1}\implies \frac{x}{(1-x)^2}=\sum_{n=0}^{\infty} nx^n$. Putting $x=\frac{1}{3}$, the answer is $\frac{\frac{1}{3}}{\frac{2}{3}\frac{2}{3}}=\frac{3}{4}$.

I'm back, and I found a reference in What is mathematics, Second edition, pg. 249.