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Calculus IIChapter 4: Series and SequencesSection 4-5

Special Series — Question 3

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Question 3

Consider ∑n=0∞12n\displaystyle\sum_{n=0}^{\infty}\frac1{2^n}.

  1. Identify the geometric data and derive the NNth partial sum.

  2. Find the sum and exact remainder.

  3. Interpret the result as repeatedly adding half of the remaining distance to 22.

Original worksheet page 1: question and worked solution for 4-5-003
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Question 3 – Solution

Step 1: Identify the series.

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

Step 2: Derive the finite partial sum.

With indices 00 through NN, there are N+1N+1 terms. The finite geometric formula gives sN=1−(1/2)N+11−1/2=2−2−N.s_N=\frac{1-(1/2)^{N+1}}{1-1/2}=2-2^{-N}. Because 2−N→02^{-N}\to0, S=limN→∞sN=2.S=\lim_{N\to\infty}s_N=2.

Step 3: Compute and interpret the remainder.

The exact gap is RN=S−sN=2−NR_N=S-s_N=2^{-N}. Starting at s0=1s_0=1, each new term equals one-half of the current distance to 22; the gap is halved at every step. The process never exceeds 22, but it can get arbitrarily close.

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