How a Telescoping Sum Collapses to Its Two Ends

Rewrite each term so almost everything cancels — only the ends survive.

Ms. Sigma

∑k=1N1k(k+1)\displaystyle\sum_{k=1}^{N} \frac{1}{k(k+1)} = N/(N+1)

  1. Split every term by partial fractions: 1/(k(k+1)) is 1/k − 1/(k+1).

    partial fractions

  2. Written out, each inside pair cancels: −½ meets +½, and −⅓ meets +⅓.

    (1 − ½) + (½ − ⅓) + (⅓ − ¼)every inside pair cancels1 − ¼

  3. Only the outermost terms are left: 1 − 1/(N + 1), which is N/(N + 1).

    only the two ends survive

How does a telescoping sum collapse to its two ends?

A telescoping sum collapses because each term splits into two pieces that cancel with its neighbors. For 1/(1 × 2) + 1/(2 × 3) + … + 1/(N(N + 1)), split each term by partial fractions: 1/(k(k + 1)) is 1/k − 1/(k + 1). Written out, the sum becomes (1 − 1/2) + (1/2 − 1/3) + (1/3 − 1/4) and so on.

Each minus 1/2 meets a plus 1/2, each minus 1/3 meets a plus 1/3, and every inside pair cancels. Only the first piece of the first term and the last piece of the last term survive, so the sum is 1 − 1/(N + 1), which is N/(N + 1). For N = 4, the sum is 1/2 + 1/6 + 1/12 + 1/20, which the formula gives as 4/5.

Check it with a common denominator of 60: 30 + 10 + 5 + 3 = 48, and 48/60 is 4/5. As N grows, the sum gets closer to 1 without reaching it.

Dr. Cipher

Your turn

Three to try in your head. Tap what you get.

Telescoping sum to 1/(12×13) = ?

Telescoping sum to 1/(6×7) = ?

Telescoping sum to 1/(10×11) = ?

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