Arithmetic series (constant difference) and geometric series (constant ratio) each have two formulas you must know by heart, plus the condition \(|r| < 1\) for a geometric series to have a sum to infinity. Recurrence sequences are examined by computing a few terms and spotting a pattern or a cycle.
Arithmetic (MM2.2)
- \(u_n = a + (n - 1)d\); \(\quad S_n = \frac n2\left[2a + (n - 1)d\right] = \frac n2(a + l)\) where \(l\) is the last term.
- \(1 + 2 + \cdots + n = \frac{n(n + 1)}{2}\).
- \(u_n = S_n - S_{n-1}\) recovers a term from a sum formula.
- Three numbers are in arithmetic progression iff the middle one is the mean of the other two.
Geometric (MM2.3)
- \(u_n = ar^{n-1}\); \(\quad S_n = \dfrac{a(r^n - 1)}{r - 1} = \dfrac{a(1 - r^n)}{1 - r}\).
- \(S_\infty = \dfrac{a}{1 - r}\) exists only when \(|r| < 1\).
- The sum of the squares of a geometric series is another geometric series with first term \(a^2\) and ratio \(r^2\).
- Three numbers \(p, q, r\) are in geometric progression iff \(q^2 = pr\).
Recurrences (MM2.1)
- \(x_{n+1} = f(x_n)\): just iterate. Look for cycles (\(x_{n+1} = \frac{1}{1 - x_n}\) has period \(3\)) and use \(n \bmod\) period for \(x_{100}\).
- To prove a sequence is increasing and bounded, show \(x_{n+1} - x_n > 0\) and use the bound in the recurrence.
- A sequence given by a formula in \(n\) may be a quadratic; its minimum is found by completing the square in \(n\) (remember \(n\) is an integer).
Must-know (no formula booklet)
- \(S_n = \frac n2[2a + (n-1)d]\)
- \(S_n = \frac{a(r^n - 1)}{r - 1}\)
- \(S_\infty = \frac{a}{1 - r}\), \(|r| < 1\)
- \(\sum_{k=1}^n k = \frac{n(n+1)}{2}\)
Classic traps
- 'Sum of the first \(n\) terms exceeds 500' needs the first \(n\) that works, so check \(n\) and \(n - 1\).
- A bouncing ball travels up and down: double the geometric series for the rises and add the initial drop.
- \(r = -1\) gives a series that does not converge even though it stays bounded.
Worked example
Q. A geometric series has \(S_\infty = 8\) and the sum of the squares of its terms is \(\frac{64}{3}\). Find \(r\).
A. \(\frac{a}{1-r} = 8\), \(\frac{a^2}{1-r^2} = \frac{64}{3}\); dividing, \(\frac{1 - r}{1 + r} = \frac13\), so \(r = \frac12\).