TheIBAxolotl
Adding forever and getting a finite answer
Take \(1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots\). Each term halves what remains to reach 2, so the running totals 1, 1.5, 1.75, 1.875, ... creep towards 2 without passing it. We say the series converges and its sum to infinity is 2.
This happens for a geometric series exactly when the terms shrink towards zero, which needs the ratio to be between \(-1\) and \(1\).
The sum to infinity
Recall \(S_n = \dfrac{u_1(1 - r^n)}{1 - r}\). If \(|r| < 1\), then \(r^n \to 0\) as \(n\) grows, and
\[S_\infty = \frac{u_1}{1 - r}, \quad |r| < 1\]
If \(|r| \ge 1\) the terms do not shrink and the series has no finite sum (it diverges).
Worked example: bouncing ball
A ball is dropped from 2 m and each bounce reaches 60% of the previous height. Find the total vertical distance travelled.
- Rebound heights: \(1.2, 0.72, 0.432, \dots\) with \(r = 0.6\); their sum is \(\dfrac{1.2}{1 - 0.6} = 3\) m.
- Each rebound is travelled up and down, giving \(2 \times 3 = 6\) m.
- Add the initial drop: total \(= 2 + 6 = 8\) m.
The key is drawing the path and noticing that the first drop is not doubled.
Recurring decimals
\(0.\overline{45} = 0.45 + 0.0045 + 0.000045 + \dots\), a geometric series with \(u_1 = 0.45\) and \(r = 0.01\):
\[0.\overline{45} = \frac{0.45}{1 - 0.01} = \frac{45}{99} = \frac{5}{11}\]
Dosage and long-term levels
A patient takes 100 mg of a medicine every day and 30% of the amount in the body remains a day later. Just after the \(n\)th dose the amount is \(100 + 100(0.3) + 100(0.3)^2 + \dots\), so in the long run it approaches \(\dfrac{100}{1 - 0.3} \approx 142.9\) mg. The same idea models repeated stocking of a lake, long-term savings with a decaying return, or total sales from a fading advertising campaign.
Finding unknowns
If you know \(S_\infty\) and one other piece of information, set up equations. For example, \(S_\infty = 20\) and \(u_1 = 5\) give \(\frac{5}{1 - r} = 20\), so \(r = 0.75\). If the ratio involves \(x\), for example \(r = \frac{x}{3}\), the condition \(|r| < 1\) gives the range \(-3 < x < 3\) for which the sum exists.
GDC tips
- Check a sum to infinity by computing \(S_n\) for a large \(n\) (say 100) with the sequence or sum function; it should be very close.
- The solver handles $\frac{u_1}{1 - r} = $ value when \(r\) is unknown.
Common mistakes
- Using the formula when \(|r| \ge 1\).
- Forgetting to double the rebounds, or doubling the first drop as well.
- Writing \(1 + r\) in the denominator.
- Not stating the convergence condition when asked to justify that the sum exists.
Exam tips
- When asked "explain why the sum to infinity exists", answer with "\(|r| = 0.6 < 1\)".
- Sketch the situation for distance problems before writing any series.
- Compare \(S_n\) with \(S_\infty\) to answer "how close" or "after how many terms" questions.