Special Series — Question 8

*Question 8 Evaluate $\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}$.

1. Differentiate the geometric power series inside its interval of convergence.

2. Align the powers and evaluate at $x=1/2$.

3. Find the finite partial sum and exact remainder.

*Question 8 – Solution

For $|x|<1$, $$\sum_{n=0}^{\infty}x^n=\frac1{1-x}.$$ Power series may be differentiated term by term at interior points, giving $$\sum_{n=1}^{\infty}nx^{n-1}=\frac1{(1-x)^2}.$$

Multiplying by $x$ yields $$\sum_{n=1}^{\infty}nx^n=\frac{x}{(1-x)^2}.$$ Since $|1/2|<1$, substitution is valid: $$\sum_{n=1}^{\infty}\frac{n}{2^n} =\frac{1/2}{(1-1/2)^2}=\boxed{2}.$$

Differentiating the finite geometric identity gives $$s_N=\sum_{n=1}^N\frac{n}{2^n}=2-\frac{N+2}{2^N}, \qquad R_N=\frac{N+2}{2^N}.$$ The remainder is positive and tends to zero, so the partial sums approach $2$ from below.