Sequences, series, and Gauss's trick
Tonight's idea
A sequence is a list of numbers with a rule. A series is what you get when you add them up. Two kinds account for most of what you will meet, each with a sum formula you can rebuild from scratch.
Arithmetic: add the same amount each time. First term \(a\), common difference \(d\), so the \(n\)th term is \(a + (n - 1)d\). Call the last term \(l\). The sum is Gauss's trick: write the series forwards and backwards and add the columns.
\[\begin{aligned} S &= a + (a + d) + \cdots + (l - d) + l \\ S &= l + (l - d) + \cdots + (a + d) + a \\ 2S &= (a + l) + (a + l) + \cdots + (a + l) \end{aligned}\]
Every column adds to \(a + l\), and there are \(n\) columns. So
\(S_n = \dfrac{n(a + l)}{2}\): number of terms times the average of first and last.
Geometric: multiply by the same amount each time. First term \(a\), common ratio \(r\), \(n\)th term \(ar^{n-1}\). Multiply the whole sum by \(r\) and subtract:
\[\begin{aligned} S &= a + ar + ar^2 + \cdots + ar^{n-1} \\ rS &= \phantom{a + {}} ar + ar^2 + \cdots + ar^{n-1} + ar^n \\ S - rS &= a - ar^n \end{aligned}\]
Everything in the middle cancels, so
\(S_n = \dfrac{a(1 - r^n)}{1 - r}\) (for \(r \ne 1\)), and if \(|r| \lt 1\), then \(r^n \to 0\) and \(S_\infty = \dfrac{a}{1 - r}\).
The famous case: \(\tfrac12 + \tfrac14 + \tfrac18 + \cdots = \dfrac{1/2}{1 - 1/2} = 1\). Not "practically 1": the sum of an infinite series is defined as the limit of the finite sums, and that limit is exactly 1.
Watch one: \(7 + 10 + 13 + \cdots + 100\), and ten doublings
Arithmetic. \(a = 7\), \(d = 3\), last term 100. How many terms? \(7 + 3(n - 1) = 100\) gives \(n = 32\). Sum: \(\dfrac{32 \times (7 + 100)}{2} = 16 \times 107 = 1712\).
Geometric. \(3 + 6 + 12 + \cdots\) for 10 terms: \(a = 3\), \(r = 2\). \(S_{10} = \dfrac{3(1 - 2^{10})}{1 - 2} = 3 \times 1023 = 3069\). (With \(r \gt 1\) it is tidier to write \(\dfrac{a(r^n - 1)}{r - 1}\). Same thing.)
Your turn
- The sequence \(4, 11, 18, 25, \ldots\) is arithmetic. Find the 50th term and the sum of the first 50 terms.
- The sequence \(5, 15, 45, \ldots\) is geometric. Find the 8th term and the sum of the first 8 terms.
- A ball is dropped from 2 m. Each bounce takes it to \(\tfrac34\) of its previous height. Find the total distance it travels before coming to rest. (Hint: the first drop is 2 m; after that every height is travelled twice, down and up.)
- Write \(0.272727\ldots\) as a geometric series and sum it to get a fraction in lowest terms. Why does this show every repeating decimal is a fraction?
Stretch: a series with a twist
Let \(S = 1 + 2x + 3x^2 + 4x^3 + \cdots\) with \(|x| \lt 1\). Compute \(S - xS\), notice what kind of series is left, sum it, and solve for \(S\). Then find \(\dfrac12 + \dfrac24 + \dfrac38 + \dfrac4{16} + \cdots\). (Hint: factor out one \(\tfrac12\) first.)
Answers
- \(d = 7\). 50th term \(= 4 + 49 \times 7 = 347\). Sum \(= \dfrac{50(4 + 347)}{2} = 25 \times 351 = 8775\).
- \(r = 3\). 8th term \(= 5 \times 3^7 = 10{,}935\). Sum \(= \dfrac{5(3^8 - 1)}{3 - 1} = \dfrac{5 \times 6560}{2} = 16{,}400\).
- First drop 2 m. Then heights \(1.5, 1.125, \ldots\), each travelled twice: \(2 \times \dfrac{1.5}{1 - 3/4} = 12\) m. Total \(2 + 12 = 14\) m. Infinitely many bounces, finite distance.
- \(0.272727\ldots = \tfrac{27}{100} + \tfrac{27}{10000} + \cdots = \dfrac{27/100}{1 - 1/100} = \dfrac{27}{99} = \dfrac{3}{11}\). Any repeating block is a geometric series with ratio \(\tfrac{1}{10^k}\), and the formula gives a fraction.
Stretch. \(xS = x + 2x^2 + 3x^3 + \cdots\). Subtracting term by term, \(S - xS = 1 + x + x^2 + \cdots = \dfrac{1}{1 - x}\). So \(S(1 - x) = \dfrac{1}{1 - x}\) and \(S = \dfrac{1}{(1 - x)^2}\). Then \(\displaystyle\sum \frac{n}{2^n} = \frac12\left(1 + 2 \cdot \tfrac12 + 3 \cdot \tfrac14 + \cdots\right) = \frac12 \cdot \frac{1}{(1 - \frac12)^2} = 2\). If you toss a fair coin until the first head, this is the average number of tosses you need.