Sum of "intertwine" of Arithmetic & Geometric Series (to infinity)

Question

I am just wondering is there any method to evaluate the following:

$$ 2\cdot (1/2) + 3\cdot (1/4) + 4\cdot (1/8) + 5\cdot (1/16) + \ldots =\sum_{n=2}^\infty \frac{n}{2^{n-1}} $$

Frankly, I have no clue how this can be solved, I have considered finding a common ratio between terms (e.g. $2\cdot (1/2)$ and $3\cdot (1/4)$ and $4\cdot (1/8))$, but no such ratio exists.

Thanks in advance.

Answer 1

$$\frac22+\frac34+\frac48+\frac5{16}+\cdots=\frac22+\frac{1+2}4+\frac{1+3}8+\frac{1+4}{16}+\cdots \\=\frac22+\frac14+\frac18+\frac1{16}+\cdots+\frac12\left(\frac22+\frac34+\frac48+\cdots\right)$$

So

$$\frac S2=\frac32.$$

Answer 2

HINT

Note that $$ (1-x)^{-1} = \sum_{k=0}^\infty x^k $$ and differentiating both sides with respect to $x$ yields $$ (1-x)^{-2} = \sum_{k=1}^\infty kx^{k-1} $$ Therefore, $$ \frac{x}{(1-x)^2} = x \sum_{k=1}^\infty kx^{k-1} = \sum_{k=1}^\infty kx^k\\ $$

Answer 3

!

I solved the problem by applying the concept used in this another question.

Concerning the current problem, by making x equal to 1/2, and factorizing x in one term, and plugging x value back in, you will get 3 as the final answer

(https://i.stack.imgur.com/tZtwG.jpg)

Answer 4

!

Concerning the current problem, by making x equal to 1/2, and factorizing x in one term, and plugging x value back in, you will get 3 as the final answer

https://i.stack.imgur.com/rhBur.jpg)