Find the sum of series $\sum_{n=1}^{\infty} \frac{2n-1}{2^n}$?

Question

Find the sum of series $\sum_{n=1}^{\infty} \frac{2n-1}{2^n}$

Answer 1

Hints:

Fill in details in the following:

$$|x|<1\implies\frac1{1-x}=\sum_{n=0}^\infty x^n\implies\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\implies\frac{2x}{(1-x)^2}=\sum_{n=1}^\infty 2nx^n$$

and now choose and sustitute in the above $\;x=...\;$ and then...

Answer 2

Hint: Use the series $$ \sum_{k=0}^\infty x^k=\frac1{1-x} $$ and $x$ times its derivative $$ \sum_{k=0}^\infty kx^k=\frac{x}{(1-x)^2} $$

Answer 3

It is clearly an absolutely convergent series, hence by setting $S=\sum_{n\geq 1}\frac{2n-1}{2^n}$ we have: $$ 2S = \sum_{n\geq 1}\frac{2n-1}{2^{n-1}}=\sum_{n\geq 1}\frac{(2(n-1)-1)+2}{2^{n-1}}=4+\sum_{n\geq 1}\frac{2(n-1)-1}{2^{n-1}} $$ or $ 2S = 3+S$, from which $\color{red}{S=3}$.