The geometric distribution, a few chapters ahead, counts trials until a first success. The fact that its probabilities sum to exactly 1 depends entirely on one infinite sum you've already met: the geometric series. This section is the one fact worth having cold before that distribution shows up.
Think of |r|<1 as each new term shrinking to a fraction of the last one. Picture a ball that bounces back to 60% of its previous height every time. It keeps bouncing forever, but the total distance it travels is still a finite number. The bounces shrink fast enough that their sum settles down instead of piling up without bound. If instead each bounce were 110% of the last, growing rather than shrinking (|r|>1), the total distance would never settle at all.
A geometric series is a sum in which each term is the previous term times a fixed constant ratio r, starting from a first term a: a,\ ar,\ ar^2,\ ar^3,\ \ldots. Written in sigma notation, \sum_{k=0}^{\infty} ar^k with |r| < 1 converges to \dfrac{a}{1-r}. When |r| \geq 1, the terms don't shrink toward zero fast enough (or they grow), and the series diverges. The sum has no finite value in that case. The finite version, \sum_{k=0}^{n-1} ar^k = a\dfrac{1-r^n}{1-r}, holds for any r \neq 1 and reduces to the infinite formula as n \to \infty exactly when |r|<1, since r^n \to 0 in that case. The ratio r can be negative, flipping the sign of every other term; the same |r|<1 test still decides convergence, since it depends only on the ratio's size, not its sign.
Formula
Sum of an Infinite Geometric Series
also known as: the convergence formula S = a_1/(1-r)
Recalled from College Algebra · Series and the Binomial Theorem — first taught there.
For |r| < 1, S = \sum_{k=1}^{\infty} a_1 r^{\,k-1} = \frac{a_1}{1 - r}. The series diverges (has no finite sum) when |r| \ge 1.
Quick review
Practice from earlier coursework, pulled in for this topic.
From Calculus II — Integral Calculus & Infinite Series
A geometric series with r=1.2...
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The sign of r plays no role -- only whether |r|<1 matters.
The formula only gives a finite, meaningful sum when |r|<1; plugging in r=1.2 anyway gives a meaningless negative number.
The size of a has no bearing on whether the series converges.
Why
Convergence of a geometric series depends only on |r|<1. With r=1.2, |r|=1.2\ge1, so the series diverges no matter what a is.
Check |r| against 1 for this ratio.
From Calculus II — Integral Calculus & Infinite Series
Evaluate \displaystyle\sum_{n=1}^\infty 8\left(\frac13\right)^{n-1}.
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Why
a=8, r=1/3, |r|<1: S=8/(1-1/3)=8/(2/3)=12.
Correct answer
Also accepted:
Why
Identify a and r, check |r|<1, then apply S = a/(1-r).
Watch a geometric series settle
Drag n up and watch the partial sum approach a fixed ceiling instead of climbing forever.
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Where the ratio determines convergence
Every ratio strictly between -1 and 1 gives a convergent series; every ratio outside that interval diverges.
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Worked example
Summing a convergent geometric series
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1
Evaluate \displaystyle\sum_{k=0}^{\infty} 3\left(\frac{1}{4}\right)^k.
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2
Here a=3 and r=\frac14, and |r|<1, so the series converges.
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3\boxed{\frac{3}{1-\frac14} = \frac{3}{\frac34} = 4}
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Worked example
Recognizing divergence
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1
Does \displaystyle\sum_{k=0}^{\infty} 2(1.2)^k converge?
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2
Here r = 1.2, and |1.2| \geq 1, so the terms grow rather than shrink.
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3\boxed{\text{Diverges — no finite sum exists.}}
Where this goes next
Chapter 5 (The geometric distribution)
A geometric random variable's probabilities are p, (1-p)p, (1-p)^2p, \ldots, a geometric series with first term p and ratio 1-p. Since 0<1-p<1 whenever 0<p<1, that series always converges, and it converges to exactly \dfrac{p}{1-(1-p)} = 1. That confirms a geometric pmf is valid before that chapter ever states the fact directly.
Check your understanding (counts toward your progress)
A geometric series has first term a=29 and sum S=62. Find the common ratio r as a number.
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Why
From S=a/(1-r): 62=29/(1-r), so 1-r=29/62, giving r=1-29/62 \approx 0.5323.
Correct answer
Also accepted:
Why
Solve S = a/(1-r) for r given the known values of S and a.
A geometric series has first term a=5 and common ratio r=0.25. What does the series converge to? Round to 2 decimal places.
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Why
a=5, r=0.25: sum = a/(1-r) = 5/0.75 ≈ 6.67.
Correct answer
Also accepted:
Why
Plug a and r directly into a/(1-r).
What is the value of \displaystyle\sum_{k=2}^{\infty} \left(\frac{1}{2}\right)^k (note: the sum starts at k=2, not k=0)?
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2 is the sum starting from k=0 — this series starts two terms later, at k=2, so it's smaller.
1.5 would be the sum starting from k=1 (missing only the k=0 term) — this series is also missing the k=1 term.
0.25 is just the FIRST term of this series (the k=2 term itself), not the total of the whole infinite series.
Why
Treat k=2 as the new starting point: the first term is a=(1/2)^2=0.25 and the ratio is still r=1/2, so the sum is a/(1-r)=0.25/0.5=0.5.
Treat k=2 as the new starting point: find the actual first term of THIS series before applying a/(1-r).
A geometric series has first term a=11 and sum S=49. Find the common ratio r as a number.
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Why
From S=a/(1-r): 49=11/(1-r), so 1-r=11/49, giving r=1-11/49 \approx 0.7755.
Correct answer
Also accepted:
Why
Solve S = a/(1-r) for r given the known values of S and a.
Practice: Applying the convergent geometric series formula (5 problems · doesn’t affect your progress)
A geometric series has ratio r = -0.8. Does it converge?
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The SIGN of r doesn't matter for convergence — only its absolute value does.
That's backwards: convergence requires |r| UNDER 1, not over 1.
Convergence depends only on r, never on the first term a's sign.
Why
|-0.8| = 0.8 < 1, so this series converges regardless of r's negative sign.
Yes — |r| = 0.8 < 1
Evaluate \displaystyle\sum_{k=0}^{\infty}\left(\frac{1}{4}\right)^k. Round to 2 decimal places.
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Why
a=1, r=1/4: sum=1/(1-1/4)=4/3≈1.33.
Correct answer
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Why
A geometric series has a=8 and r=0.5. What is the sum?
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Why
a/(1-r) = 8/0.5 = 16.
Correct answer
Also accepted:
Why
Plug a and r directly into a/(1-r).
A geometric series with first term a=4 converges to 10. What is the common ratio r?
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Why
Solve 4/(1-r)=10: 1-r=0.4, so r=0.6.
Correct answer
Also accepted:
Why
Set up a/(1-r)=10 with a=4, then solve for r algebraically.
A geometric series begins \displaystyle\sum_{k=3}^{\infty} 2\left(\frac{1}{3}\right)^k (note the starting index is k=3). Evaluate the sum as a fraction in lowest terms.
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Any equivalent form counts — 2x+3, 3 + 2·x, … Use the keypad below the field for fractions, roots, and other symbols.
Why
The series' true first term is at k=3: a=2(1/3)^3=2/27. With r=1/3: sum =\dfrac{2/27}{1-1/3}=\dfrac{2/27}{2/3}=\dfrac19.
Correct answer
Also accepted:
Why
Find the actual FIRST term of the series — the one at k=3, not k=0 — then apply a/(1-r) using that as a.
a = 2(1/3)^3 = 2/27.