Evaluate the sum of 1/(n(n+1)) from n=1 to infinity: animated lesson

The problem. Here is the problem: Σ 1 / (n times (n plus 1)), n equals 1 to ∞. Decide whether Σ 1 / (n(n plus 1)) (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 1 / (n(n plus 1)). First terms: 0.5, 0.166667, 0.0833333, 0.05, …. nth-term test. lim a(n) equals 0, so the divergence test is inconclusive; the terms do shrink, so a finer test is needed. Limit comparison with a p-series. Positive terms: the terms shrink roughly like 1/n squared (estimated from a(2N)/a(N)), so compare with Σ 1/n squared: lim a(n) times n squared equals 1. The limit is a positive finite number, so the series behaves exactly like the p-series Σ 1/n squared, which converges because p greater than 1. 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 1. Any closed form shown is a numerical identification, not a proof.

The answer. So the answer is The series converges (limit comparison with Σ 1/n squared); sum approximately 1 (matches 1 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 0.90909091, S₁₀₀ equals 0.99009901, S₁₀₀₀ equals 0.999001, S₁₀₀₀₀ equals 0.99990001.