Telescoping Series

Question

Find the sum of the telescoping series whose general term is 1/(k(k+1)) summed from k equal to 1 up to k equal to n.

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A telescoping series is summed by writing each term as a difference of consecutive quantities so that interior terms cancel in pairs, a method central to JEE Advanced summation. Using partial fractions, 1/(k(k+1)) = 1/k - 1/(k+1). Summing from k = 1 to n produces (1 - 1/2) + (1/2 - 1/3) + ... + (1/n - 1/(n+1)), and every intermediate fraction cancels, leaving 1 - 1/(n+1). Combining gives (n+1-1)/(n+1) = n/(n+1). Option 1/(n+1) keeps the tail rather than the surviving head minus tail. Option (n-1)/n misindexes the cancellation boundary. Option n/(n-1) inverts the structure entirely. Hence the sum is n/(n+1). The decisive insight is that telescoping works only because the partial-fraction pieces are consecutive, so each negative fragment of one term is annihilated by the positive fragment of the next, leaving an exact closed form rather than an approximation. Plausibility check: as n grows large the sum approaches 1, which agrees with the known convergent value of the full infinite series, and for n = 1 the formula gives 1/2 = 1/(1·2), matching the single term and the position-one boundary exactly.

Telescoping Series

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