Sums that collapse
Tonight's idea
Add these: \(\dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dfrac{1}{3 \times 4} + \dots + \dfrac{1}{99 \times 100}\). Ninety-nine fractions, and a common denominator would be enormous. Look at one term instead. \(\tfrac12 - \tfrac13 = \tfrac{3 - 2}{6} = \tfrac16\), which is \(\tfrac{1}{2 \times 3}\). The same thing happens every time:
\[\frac1n - \frac1{n+1} = \frac{(n+1) - n}{n(n+1)} = \frac{1}{n(n+1)}.\]So rewrite every term as a difference:
\[\left(1 - \tfrac12\right) + \left(\tfrac12 - \tfrac13\right) + \left(\tfrac13 - \tfrac14\right) + \dots + \left(\tfrac1{99} - \tfrac1{100}\right).\]The \(-\tfrac12\) meets a \(+\tfrac12\). The \(-\tfrac13\) meets a \(+\tfrac13\). Everything in the middle cancels. Only the two ends survive: \(1 - \tfrac1{100} = \tfrac{99}{100}\). This is a telescoping sum, named after a spyglass that slides shut until only its ends show.
If each term is (one thing) − (the next thing), the whole sum is first − last.
Watch one: a gap of 2
Find \(\dfrac{1}{1 \times 3} + \dfrac{1}{3 \times 5} + \dfrac{1}{5 \times 7} + \dots + \dfrac{1}{99 \times 101}\). Try the same split: \(1 - \tfrac13 = \tfrac23\). That is twice \(\tfrac{1}{1 \times 3}\), not once. In general \(\dfrac1n - \dfrac1{n+2} = \dfrac{2}{n(n+2)}\), so each term is half a difference:
\[\tfrac12\left[\left(1 - \tfrac13\right) + \left(\tfrac13 - \tfrac15\right) + \dots + \left(\tfrac1{99} - \tfrac1{101}\right)\right] = \tfrac12\left(1 - \tfrac1{101}\right) = \tfrac{50}{101}.\]Next year: what happens when the sum never stops, in Adding up forever.
Your turn
- (a) Check by subtracting that \(\tfrac13 - \tfrac14 = \tfrac1{12}\) and \(\tfrac14 - \tfrac15 = \tfrac1{20}\). (b) Find \(\dfrac{1}{1 \times 2} + \dfrac{1}{2 \times 3} + \dots + \dfrac{1}{9 \times 10}\) as a single fraction.
- This sum starts in the middle. Find \(\dfrac{1}{10 \times 11} + \dfrac{1}{11 \times 12} + \dots + \dfrac{1}{19 \times 20}\).
- Products can collapse too. (a) Write \(1 - \tfrac12\), \(1 - \tfrac13\) and \(1 - \tfrac14\) as single fractions. (b) Find \(\left(1 - \tfrac12\right)\left(1 - \tfrac13\right)\left(1 - \tfrac14\right) \cdots \left(1 - \tfrac1{100}\right)\).
- (a) Write \(\dfrac1n - \dfrac1{n+3}\) as a single fraction. (b) Use it to find \(\dfrac{1}{1 \times 4} + \dfrac{1}{4 \times 7} + \dfrac{1}{7 \times 10} + \dots + \dfrac{1}{28 \times 31}\).
Stretch: square roots that collapse
Expand \(\left(\sqrt{n+1} + \sqrt{n}\right)\left(\sqrt{n+1} - \sqrt{n}\right)\). Use the result to write \(\dfrac{1}{\sqrt{n} + \sqrt{n+1}}\) as a difference, then find
\[\frac{1}{\sqrt1 + \sqrt2} + \frac{1}{\sqrt2 + \sqrt3} + \dots + \frac{1}{\sqrt{99} + \sqrt{100}}.\]Answers
- (a) \(\tfrac13 - \tfrac14 = \tfrac{4 - 3}{12} = \tfrac1{12}\) and \(\tfrac14 - \tfrac15 = \tfrac{5 - 4}{20} = \tfrac1{20}\). (b) \(\left(1 - \tfrac12\right) + \left(\tfrac12 - \tfrac13\right) + \dots + \left(\tfrac19 - \tfrac1{10}\right) = 1 - \tfrac1{10} = \tfrac9{10}\).
- \(\left(\tfrac1{10} - \tfrac1{11}\right) + \left(\tfrac1{11} - \tfrac1{12}\right) + \dots + \left(\tfrac1{19} - \tfrac1{20}\right) = \tfrac1{10} - \tfrac1{20} = \tfrac1{20}\). First minus last still works; the first is now \(\tfrac1{10}\).
- (a) \(\tfrac12\), \(\tfrac23\), \(\tfrac34\). (b) \(\tfrac12 \times \tfrac23 \times \tfrac34 \times \dots \times \tfrac{99}{100}\). Each numerator cancels the denominator before it, leaving \(\tfrac1{100}\).
- (a) \(\dfrac{(n + 3) - n}{n(n + 3)} = \dfrac{3}{n(n + 3)}\). (b) Each term is a third of a difference, so the sum is \(\tfrac13\left[\left(1 - \tfrac14\right) + \left(\tfrac14 - \tfrac17\right) + \dots + \left(\tfrac1{28} - \tfrac1{31}\right)\right] = \tfrac13\left(1 - \tfrac1{31}\right) = \tfrac13 \times \tfrac{30}{31} = \tfrac{10}{31}\).
Stretch. By the difference of two squares the product is \((n + 1) - n = 1\). So \(\dfrac{1}{\sqrt{n} + \sqrt{n+1}} = \sqrt{n+1} - \sqrt{n}\). The sum is \(\left(\sqrt2 - \sqrt1\right) + \left(\sqrt3 - \sqrt2\right) + \dots + \left(\sqrt{100} - \sqrt{99}\right) = \sqrt{100} - \sqrt1 = 10 - 1 = 9\).