Why an Infinite Sum Can Still Add Up to a Finite Number

When each term is multiplied by a fixed number instead of increased by one — a loan growing by compound interest, a ball bouncing to half its height — the sequence is a geometric progression. GPs grow or shrink far faster than APs, and some infinite GPs even have a finite sum.

This part covers GP terms and sums, geometric means and their link with arithmetic means, sums of powers of natural numbers, and the method of differences.

**A GP with first term a and common ratio r has nth term $a_n=a{r}^{n-1}$, sum $S_n=\frac{a(r^n-1)}{r-1}$ for $r\ne 1$, and, when $|r|<1$, sum to infinity $S_{\infty }=\frac{a}{1-r}$.

Deriving the sum.** Multiply $S_n=a+ar+\cdots +a{r}^{n-1}$ by r and subtract: $r{S}_n-S_n=a{r}^n-a$, which gives the formula. When $|r|<1$, $r^n$ shrinks towards 0 as n grows, leaving $\frac{a}{1-r}$.

Worked example. For $3,6,12,\dots$: $a_8=3\times 2^7=384$ and $S_8=\frac{3(2^8-1)}{2-1}=765$.

Worked example 2. The recurring decimal $0.777\dots =0.7+0.07+0.007+\cdots =\frac{0.7}{1-0.1}=\frac{7}{9}$.

An everyday example. A ball dropped from 10 m that bounces to half its height each time travels $10+2\times \frac{5}{1-0.5}=30$ m in total before coming to rest.

The substance. **An infinite GP has a sum only when $|r|<1$** — for $r=2$ the terms grow without limit.

**The geometric mean of positive numbers a and b is $G=\sqrt{ab}$, inserting n geometric means between a and b gives a GP with common ratio $r={\left(\frac{b}{a}\right)}^{\frac{1}{n+1}}$, and for positive numbers the arithmetic mean is never less than the geometric mean.

Worked example.** Insert 3 geometric means between 2 and 162.

$$r={\left(\frac{162}{2}\right)}^{\frac{1}{4}}=81^{\frac{1}{4}}=3$$

The means are 6, 18 and 54, giving 2, 6, 18, 54, 162.

AM–GM inequality. For positive a and b,

$$A-G=\frac{a+b}{2}-\sqrt{ab}=\frac{(\sqrt{a}-\sqrt{b}{)}^2}{2}\ge 0$$

with equality only when $a=b$. For 4 and 9, $A=6.5$ and $G=6$.

An everyday example. Of all rectangular plots with the same perimeter, the square encloses the largest area — a direct consequence of $A\ge G$ applied to the length and breadth.

The substance. **Inserting n means creates $n+1$ ratios** — so the power is $\frac{1}{n+1}$, exactly as with arithmetic means.

**The sums of the first n natural numbers, their squares and their cubes are $\sum k=\frac{n(n+1)}{2}$, $\sum k^2=\frac{n(n+1)(2n+1)}{6}$ and $\sum k^3={\left[\frac{n(n+1)}{2}\right]}^2$, and a series with a polynomial kth term is summed by splitting it into these three.

Worked example.** $1^2+2^2+\cdots +10^2=\frac{10\times 11\times 21}{6}=385$.

Worked example 2. Sum n terms of $1\times 2+2\times 3+3\times 4+\cdots$. The kth term is $k(k+1)=k^2+k$, so

$$S_n=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$$

For $n=4$: $2+6+12+20=40$, and $\frac{4\times 5\times 6}{3}=40$.

A neat link. $\sum k^3={\left(\sum k\right)}^2$; for $n=3$, $1+8+27=36=6^2$.

An everyday example. Oranges stacked on a fruit cart as a square pyramid of 6 layers number $1+4+9+16+25+36=\frac{6\times 7\times 13}{6}=91$.

The substance. Write the kth term first — the formulae apply to k, not directly to the numbers printed in the question.

**When the differences between consecutive terms form an AP or a GP, the nth term is the first term plus the sum of the first $n-1$ differences, found with the AP or GP sum formula.

Worked example.** Find the nth term of $2,5,10,17,26,\dots$

- Differences: $3,5,7,9,\dots$, an AP with first term 3 and difference 2
- Sum of the first $n-1$ differences: $\frac{n-1}{2}[6+2(n-2)]=(n-1)(n+1)=n^2-1$
- So $a_n=2+n^2-1=n^2+1$, and indeed $a_5=26$
- The sum of n terms is then $\frac{n(n+1)(2n+1)}{6}+n$; for $n=3$ this gives $14+3=17=2+5+10$

Worked example 2. For $1,3,7,15,31,\dots$ the differences $2,4,8,16$ form a GP, so $a_n=1+(2+4+\cdots +2^{n-1})=2^n-1$.

An everyday example. Matchsticks for triangular grid patterns go 3, 9, 18, 30; the differences 6, 9, 12 form an AP, and the method gives $\frac{3n(n+1)}{2}$ sticks for the nth pattern.

The substance. If the first differences show no pattern, try the second differences — a constant second difference means the nth term is quadratic in n.