How do I evaluate $\sum\limits_{n=1}^\infty(n-1)\cdot\big(\frac12\big)^n$?

Question

I know from the ratio test that the series converges and I used wolfram alpha to determine that the answer is 1, but what are the steps involved to reach this answer?

Answer 1

$$\sum\limits_{n=1}^\infty(n-1)\cdot\big(\frac12\big)^n=\frac12\sum\limits_{n=1}^\infty n\big(\frac12\big)^n$$$$=\frac12\sum\limits_{n=1}^\infty\sum\limits_{m=n}^\infty\big(\frac12\big)^m$$$$=\frac12\sum\limits_{n=1}^\infty\big(\frac12\big)^{n-1}$$$$=\sum\limits_{n=1}^\infty\big(\frac12\big)^n$$$$=\color{red}1$$

Answer 2

Start with a geometric series \begin{eqnarray*} \sum_{n=0}^{\infty} x^n = \frac{1}{1-x}. \end{eqnarray*} Differentiate \begin{eqnarray*} \sum_{n=0}^{\infty} n x^{n-1} = \frac{1}{(1-x)^2}. \end{eqnarray*} Multiply by $x^2$ and change the summation index \begin{eqnarray*} \sum_{n=1}^{\infty} (n-1) x^{n} = \frac{x^2}{(1-x)^2}. \end{eqnarray*} Now just substitute $x=1/2$ and you will have the result.

Answer 3

Hint Use the geometric series $$1+x+x^2+\dotsb + =\frac{1}{1-x},$$ and then take the derivative to obtain $$x(1+2x+3x^2+\dotsb+)=x\left(\frac{1}{(1-x)^2}\right)$$

Answer 4

First $\sum_{n\ge0}x^n=\dfrac1{1-x}$ for $\mid x\mid\lt1$.

And differentiating, we get $\sum_{n\ge0}n(1/2)^{n-1}=\dfrac 1{(1-1/2)^2}=4$.

So, we get $\sum_{n\ge0}(n-1)(1/2)^n=(1/4)(\sum_{n\ge2} (n-1)(1/2)^{n-2})=1$.