The sum 1/(√1 + √2) + 1/(√2 + √3) + 1/(√3 + √4) + … + 1/(√99 + √100) equals:

Rationalising a single term, 1/(√k + √(k + 1)) = (√(k + 1) − √k)/((k + 1) − k) = √(k + 1) − √k. The sum therefore telescopes: (√2 − √1) + (√3 − √2) + … + (√100 − √99) = √100 − √1 = 10 − 1 = 9. (10 forgets the leftover −√1; 99 is merely the number of terms.) The first three terms alone give √4 − 1 = 1, which matches 0.414 + 0.318 + 0.268.