Multiply instead of add, and a series either explodes to infinity or — remarkably — settles on a single finite limit.
10 min read4 worked examples
The nth term of a geometric sequence
A geometric sequence is a sequence in which each term is found by multiplying the previous one by a fixed common ratio $r$. Starting from a first term $a$, the sequence runs $a,\ ar,\ ar^2,\ ar^3,\dots$, so the $n$th term is
$u_n = ar^{n-1}$
Each term picks up one more factor of $r$ than the last, and $(n-1)$ counts how many multiplications have been applied since the first term. If $r>1$ the terms grow without bound; if $0<r<1$ they shrink towards zero; if $r<0$ the terms alternate in sign.
The sum of a geometric series
For a geometric series (the sum of the terms of a geometric sequence), the sum of the first $n$ terms is given by
$S_n = \frac{a(1-r^n)}{1-r}, \qquad r\ne1$
ProofThe sum formula, by multiplying and subtracting
1
Write $S_n$ as the sum of $n$ terms: $S_n = a+ar+ar^2+\cdots+ar^{n-1}$.
Direct definition of the sum of $n$ terms.
2
Multiply every term of $S_n$ by $r$: $rS_n = ar+ar^2+ar^3+\cdots+ar^{n}$.
Shifts every term of the original series up by one power of $r$.
3
Subtract: $S_n - rS_n = a - ar^n$, since every middle term ($ar, ar^2,\dots,ar^{n-1}$) appears in both sums and cancels.
Only the first term of $S_n$ and the last term of $rS_n$ survive the subtraction.
4
Factorise both sides: $S_n(1-r) = a(1-r^n)$.
Take out the common factor $S_n$ on the left and $a$ on the right.
5
Divide by $(1-r)$, valid since $r\ne1$: $S_n = \dfrac{a(1-r^n)}{1-r}$.
Isolates $S_n$, giving the standard sum formula.
∎ proven
Sum to infinity
If $|r|<1$, then as $n\to\infty$, $r^n\to0$, so the finite sum $S_n$ settles on a limiting value called the sum to infinity:
$S_\infty = \frac{a}{1-r}, \qquad |r|<1$
ProofThe sum-to-infinity formula, as the limit of $S_n$
1
Start from the finite sum formula, valid for $r\ne1$: $S_n = \dfrac{a(1-r^n)}{1-r}$.
Already proven above for any number of terms $n$.
2
Consider what happens to $r^n$ as $n\to\infty$, given that $|r|<1$.
Repeatedly multiplying a number of magnitude less than 1 by itself makes it shrink steadily towards zero.
3
Since $|r|<1$, $r^n\to0$ as $n\to\infty$, so $S_n \to \dfrac{a(1-0)}{1-r} = \dfrac{a}{1-r}$.
Substitute the limiting value $r^n=0$ into the finite sum formula.
4
This limiting value is defined to be the sum to infinity: $S_\infty = \dfrac{a}{1-r}$.
$S_\infty$ is, by definition, the value that $S_n$ approaches as $n\to\infty$.
∎ proven
A series with this property is called convergent. The modulus condition $|r|<1$ means $-1<r<1$: it captures both positive ratios shrinking towards zero ($0<r<1$) and alternating ratios shrinking in magnitude ($-1<r<0$). If $|r|\ge1$, the terms never shrink — $r^n$ either grows unboundedly or fails to settle — so the series is divergent and has no sum to infinity.
Why the modulus condition matters
Writing $|r|<1$ rather than just $r<1$ is essential: a ratio like $r=-0.9$ satisfies $|r|<1$ and gives a perfectly convergent series (terms alternate in sign but shrink towards zero), whereas $r=-2$ has $r<1$ true but the series clearly diverges, since $|{-2}|=2>1$ means the terms grow without bound in size. The modulus is what correctly captures "shrinking towards zero," regardless of sign.
Using logarithms to find n
Because $n$ appears as an exponent in $u_n=ar^{n-1}$ and $S_n=\frac{a(1-r^n)}{1-r}$, solving for $n$ given a target value requires taking logarithms of both sides to bring the exponent down, then rearranging.
Worked example 1
Finding a term from the nth-term formula
A geometric sequence has first term $a=3$ and common ratio $r=2$. Find the 8th term.
MethodSubstitute directly into $u_n=ar^{n-1}$
1
$u_8 = ar^{8-1} = 3(2)^7$.
Substitute $n=8$, $a=3$, $r=2$ into the nth-term formula.
2
$u_8 = 3(128) = 384$.
Evaluate $2^7=128$ and multiply.
▪ solved
Examiner's eye
Since $r=2>1$, the sequence grows rapidly — a useful sanity check that the answer should be considerably larger than $a$.
Worked example 2
Sum to infinity of a convergent series
A geometric series has first term $a=12$ and common ratio $r=\dfrac13$. Show the series converges, and find its sum to infinity.
MethodCheck $|r|<1$, then apply $S_\infty=\frac{a}{1-r}$
1
$\left|\dfrac13\right| = \dfrac13 < 1$, so the series converges.
Substitute $a=12$, $r=\frac13$ into the sum-to-infinity formula.
3
$S_\infty = 12\times\dfrac32 = 18$.
Divide by a fraction by multiplying by its reciprocal.
▪ solved
Strategy
Always state the convergence check $|r|<1$ explicitly before quoting $S_\infty$ — in an exam, this is often a method mark in its own right, not just a formality.
Worked example 3
Using logarithms to find n
A geometric sequence has first term $a=5$ and common ratio $r=1.2$. Find the least value of $n$ for which $u_n > 100$.
MethodSet up an inequality, take logs of both sides, then solve for n
Substitute into $u_n=ar^{n-1}$ and divide both sides by 5.
2
Take logarithms of both sides: $\log\big((1.2)^{n-1}\big) > \log(20) \Rightarrow (n-1)\log(1.2) > \log(20)$.
Logarithms allow the exponent $(n-1)$ to be brought down as a multiplier.
3
Since $\log(1.2)>0$, divide both sides by $\log(1.2)$ without flipping the inequality: $n-1 > \dfrac{\log(20)}{\log(1.2)} \approx \dfrac{1.301}{0.0792} \approx 16.43$.
Dividing by a positive number preserves the direction of the inequality.
4
$n > 17.43$, so the least integer value is $n=18$.
$n$ must be a whole number, and must exceed 17.43, so round up.
▪ solved
The key insight
When the common ratio $r$ satisfies $0<r<1$, $\log(r)$ is negative — dividing an inequality by a negative number flips its direction. Always check the sign of $\log(r)$ before dividing, or the final inequality will point the wrong way.
Worked example 4
Finding a and r from given terms, then the sum to infinity · Challenge
A geometric series has $u_2=6$ and $u_4=\dfrac{3}{2}$. Given that the series converges, find $a$, $r$, and $S_\infty$.
MethodDivide successive nth-term equations to find r, then back-substitute
Dividing eliminates $a$, leaving an equation purely in $r$.
3
$r = \pm\dfrac12$. Since the series converges, $|r|<1$ — both values satisfy this, so both are valid candidates pending the sign check.
Square roots give two possible ratios; the convergence condition alone doesn't distinguish them.
4
If $r=\dfrac12$: $a(\tfrac12)=6 \Rightarrow a=12$. If $r=-\dfrac12$: $a(-\tfrac12)=6 \Rightarrow a=-12$.
Back-substitute each candidate $r$ into $ar=6$.
5
Both are self-consistent (check $u_4$: $12(\tfrac12)^3=\tfrac32$ ✓, and $-12(-\tfrac12)^3 = -12(-\tfrac18)=\tfrac32$ ✓), so both are genuine solutions.
Verify each pair $(a,r)$ against both original conditions.
6
For $(a,r)=(12,\tfrac12)$: $S_\infty=\dfrac{12}{1-\frac12}=24$. For $(a,r)=(-12,-\tfrac12)$: $S_\infty=\dfrac{-12}{1-(-\frac12)}=\dfrac{-12}{3/2}=-8$.
Apply $S_\infty=\frac{a}{1-r}$ to each valid pair.
▪ solved
Look deeper — square roots often give two valid branches
Unlike a squaring-a-modulus-equation situation, here both values of $r$ genuinely satisfy every given condition — there is no extraneous root to discard. A question would need extra information (e.g. "the terms are all positive") to select a single answer; otherwise both $(a,r)$ pairs and both sums to infinity are valid.
Common mistake
Quoting a sum to infinity without checking $|r|<1$ first — if $|r|\ge1$ the series diverges and $S_\infty$ simply does not exist, no matter what the formula $\frac{a}{1-r}$ might numerically evaluate to. Also watch for sign errors when $r$ is negative: $1-r$ becomes $1-(\text{negative})=1+|r|$, not $1-|r|$.
Check yourself
A geometric series has first term $a=8$ and common ratio $r=-\dfrac14$. Show the series converges and find $S_\infty$. (Hint: check $|r|<1$ first, being careful with the negative sign when substituting into the formula.)
Method
1
$\left|-\dfrac14\right| = \dfrac14 < 1$, so the series converges.