Special Series — Question 3

*Question 3 Consider $\displaystyle\sum_{n=0}^{\infty}\frac1{2^n}$.

1. Identify the geometric data and derive the $N$th partial sum.

2. Find the sum and exact remainder.

3. Interpret the result as repeatedly adding half of the remaining distance to $2$.

*Question 3 – Solution

The expansion is $1+1/2+1/4+\cdots$, so the first term is $a=1$ and the common ratio is $r=1/2$. Since $|r|<1$, the series converges.

With indices $0$ through $N$, there are $N+1$ terms. The finite geometric formula gives $$s_N=\frac{1-(1/2)^{N+1}}{1-1/2}=2-2^{-N}.$$ Because $2^{-N}\to0$, $$S=\lim_{N\to\infty}s_N=2.$$

The exact gap is $R_N=S-s_N=2^{-N}$. Starting at $s_0=1$, each new term equals one-half of the current distance to $2$; the gap is halved at every step. The process never exceeds $2$, but it can get arbitrarily close.