Summation of series
Standard summation results
$$\sum_{r=1}^{n} r = \tfrac12 n(n+1), \qquad \sum_{r=1}^{n} r^2 = \tfrac16 n(n+1)(2n+1),$$
$$\sum_{r=1}^{n} r^3 = \tfrac14 n^2(n+1)^2.$$
- Split a sum into these standard pieces, e.g. $\sum(2r+1) = 2\sum r + \sum 1 = n(n+2)$.
Practice
Using Σr = ½n(n+1), what is the sum of the first 5 integers?
½ × 5 × 6 = 15 (= 1+2+3+4+5).
Practice
Using Σr² = (1/6)n(n+1)(2n+1), what is the sum of the first 3 squares?
(1/6)(3)(4)(7) = 84/6 = 14 (= 1+4+9).
The method of differences
- If each term is $f(r) - f(r+1)$, almost everything cancels (telescopes).
- From the sum to $n$ terms you can see if a series converges and find its sum to infinity.
Practice
In the method of differences, if each term is f(r) − f(r+1), most terms cancel (telescope).
Consecutive terms cancel, leaving only the first and last pieces.
You've got it
Key idea
- $\sum r = \tfrac12 n(n+1)$, $\sum r^2 = \tfrac16 n(n+1)(2n+1)$, $\sum r^3 = \tfrac14 n^2(n+1)^2$
- split sums into standard pieces
- method of differences telescopes $f(r) - f(r+1)$