Sum 1/(k(k+2)) for k running from 1 to 20, as an exact fraction. What does the sum approach as the upper limit grows?

Sum 1/(k(k+2)) for k running from 1 to 20, as an exact fraction. What does the sum approach as the upper limit grows?

Approach: Split the term into partial fractions so consecutive terms cancel, then track which boundary terms survive at each end given the gap of 2 in the denominators.

325/462. Partial fractions give 1/(k(k+2)) = (1/2)(1/k - 1/(k+2)), so the telescoping sum leaves only the first two positive terms and the last two negative ones: (1/2)(1 + 1/2 - 1/21 - 1/22). That works out as (1/2)(3/2 - 43/462) = (1/2)(650/462) = 325/462 = 0.70346. The gap of 2 in the denominators is why two terms survive at each end rather than one, which is the step candidates get wrong most often. As the upper limit grows the limit of the sum approaches 3/4, since (1/2)(1 + 1/2) is 3/4. The remaining error is (1/2)(1/(n+1) + 1/(n+2)), which at an upper limit of 20 is 0.04654 and matches 0.75 - 0.70346 exactly.

Follow-up: What is the sum of 1/(k(k+3)) from k = 1 to infinity, and how many boundary terms survive there?

Key concepts: telescoping sum, partial fractions, limit of the sum.