The Collapsing Sum

Question

A contract pays frac1n(n+1) dollars on day n, for n = 1, 2, ldots, 99. Ignoring discounting, compute the total payout exactly: no term-by-term grinding allowed.

Accepted Answer

Partial fractions. Since frac1n(n+1) = frac1n - frac1n+1 (verify: common denominator gives frac(n+1)-nn(n+1)), the sum telescopes: sum_n=1^99left(frac1n - frac1n+1right) = left(1 - tfrac12right) + left(tfrac12 - tfrac13right) + cdots + left(tfrac199 - tfrac1100right) = 1 - frac1100 = mathbf0.99. Every interior term appears once with a plus and once with a minus; only the endpoints survive. In general sum_n=1^N frac1n(n+1) = fracNN+1 to 1. Why a trader should care about the move, not just the answer. Telescoping is "decompose into adjacent differences", the same trick as a strip of calendar spreads: buy every spread between month 1 and month 100 and your position nets to front month minus back month. Whenever a sum collapses, look for the difference representation first. A probability bonus that interviewers love: because the infinite sum equals 1, the numbers frac1n(n+1) form a probability distribution. It is exactly the law of "how long does the current best stay the best": for iid continuous draws, the chance the first draw still leads after n more is frac1n+1, so P(textdethroned at step n) = frac1n - frac1n+1. The mean of that distribution is the harmonic series, infinite. A rec

The Collapsing Sum

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