1.8 Infinite geometric seriesIB Maths: Analysis and Approaches HL: Revision notes

The sum of the first $n$ terms of a geometric series is $S_n = \frac{u_1(1-r^n)}{1-r}$. If $|r|<1$, then $r^n \to 0$ as $n\to\infty$, so $S_n$ gets closer and closer to a fixed value. The series is then convergent.

If $|r|\ge1$ the terms do not shrink to zero and the series has no sum to infinity — it is divergent. The condition for convergence is written with modulus notation: $|r|<1$, i.e. $-1<r<1$.

For a convergent geometric series,$S_\infty = \frac{u_1}{1-r}, \quad |r|<1.$This is in the formula booklet. Example: $u_1 = 24$, $r = -\frac12$ gives $S_\infty = \frac{24}{1+\frac12} = 16$.

A negative ratio makes the partial sums oscillate above and below the limit: $24, 12, 18, 15, \ldots \to 16$.

When the ratio contains a variable, convergence gives an inequality to solve. For $1 + (2x-1) + (2x-1)^2 + \cdots$, $r = 2x-1$, so$|2x-1|<1 \iff -1<2x-1<1 \iff 0<x<1.$Inside this interval, $S_\infty = \frac{1}{1-(2x-1)} = \frac{1}{2-2x}$. After solving for $x$ from a given sum, always check that your value lies in the interval of convergence.

A ball dropped from 3 m that rebounds to 60% of each height travels $3$ m down, then $1.8$ m up and $1.8$ m down, then $1.08$ m up and down, and so on:$3 + 2\times\frac{1.8}{1-0.6} = 3 + 9 = 12\text{ m}.$Watch which distances are travelled twice. The model is idealised: a real ball stops after finitely many bounces, but the infinite sum gives a good upper bound.

If you are told $S_\infty$ and another fact (such as the sum of the first two terms), write two equations and eliminate $u_1$. From $\frac{a}{1-r} = 45$ and $a(1+r) = 40$: $45(1-r)(1+r) = 40$, so $1-r^2 = \frac89$ and $r = \pm\frac13$. Both values give a valid series because $|r|<1$; extra information (e.g. all terms positive) chooses between them.

Every partial sum of a series with positive terms is less than $S_\infty$, because infinitely many positive terms are still to be added.