Series - The Basics — Question 7

*Question 7 Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}$.

1. Differentiate the geometric-series identity to derive a formula for $\sum_{n=1}^{\infty}nx^{n-1}$ when $|x|<1$.

2. Use that formula at $x=1/2$ to evaluate the series.

3. Derive the finite partial sum and exact remainder after $N$ terms.

*Question 7 – Solution

For $|x|<1$, the geometric identity is $$\sum_{n=0}^{\infty}x^n=\frac1{1-x}.$$

Power series may be differentiated term by term at every interior point of their interval of convergence. Thus $$\sum_{n=1}^{\infty}nx^{n-1}=\frac1{(1-x)^2}.$$ Multiplying both sides by $x$ aligns the power with the desired form: $$\sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}.$$

Since $|1/2|<1$, substitution is valid and gives $(1/2)/(1/2)^2=2$.

Differentiating the finite identity $1+x+\cdots+x^N=(1-x^{N+1})/(1-x)$ and multiplying by $x$ yields $$\sum_{n=1}^N nx^n=\frac{x-(N+1)x^{N+1}+Nx^{N+2}}{(1-x)^2}.$$ At $x=1/2$, this simplifies to $$s_N=2-\frac{N+2}{2^N},\qquad R_N=2-s_N=\frac{N+2}{2^N}.$$