Evaluate the sum of n/2^n from n=1 to infinity: animated lesson

The problem. Here is the problem: Σ n / 2 ^ n, n equals 1 to ∞. Decide whether Σ n / 2 to the power n (n from 1 to ∞) converges or diverges.

Step by step. Write the series. Σ a(n) for n equals 1, 2, 3, … with a(n) equals n / 2 to the power n. First terms: 0.5, 0.5, 0.375, 0.25, …. nth-term test. lim a(n) equals 0, so the divergence test is inconclusive; the terms do shrink, so a finer test is needed. Ratio test. Positive terms: a(n plus 1)/a(n) simplifies to (n plus 1) / 2 / n, and its limit as n → ∞ is L equals 1 over 2. L less than 1, so the series converges (absolutely). Estimate the sum numerically. Partial sums S N were computed for several N and extrapolated to N → ∞ (Richardson extrapolation; alternating terms are averaged first). Two independent extrapolations agree to better than 10 to the power minus 10, giving S approximately 2. Any closed form shown is a numerical identification, not a proof.

The answer. So the answer is The series converges (ratio test); sum approximately 2 (matches 2 to 10+ digits). A series converges when its partial sums approach a finite number. The terms must tend to 0 for that to happen, but that alone is not enough (the harmonic series diverges), so a comparison, ratio, root, alternating or integral test is used to decide.

Check. Partial sums (supporting evidence, not a proof): S₁₀ equals 1.9882813, S₁₀₀ equals 2, S₁₀₀₀ equals 2, S₁₀₀₀₀ equals 2.