A series is what you get by adding the terms of a sequence one after another, starting from the first.
Whether the running total settles at one value or grows forever depends on how quickly the terms shrink.
A geometric series adds up to a single value only when its ratio is less than 1 in absolute value; otherwise it diverges.
Move the ratio r and the number of terms n to see whether the partial-sum bars approach the dashed line at 1/(1 − r) or keep growing.
Adding Forever
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Partial sums Sₙ of the geometric series 1 + r + r² + … and the limit 1/(1 − r)
8
0.5
👀 What to read in the picture
①For the geometric series with first term 1, the n-th bar is Sₙ, the sum of the first n terms
②If |r| < 1, the bars approach the red dashed line 1/(1 − r); for r = 0.5 that value is 2
③If r is negative, the bars approach it while jumping above and below the line; if r ≥ 1, they grow without bound and the series diverges
Judging by the Size of the Terms
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Each term's size |r|ᵏ drawn as a square; it changes with the ratio slider above
📦 Convergence in squares
①Each square is one term of the series, and its side is proportional to the size of that term, |r|ᵏ
②If |r| < 1, the squares keep shrinking by the same factor, so even all of them together add up to a finite amount
③If |r| = 1 the size stays the same, and if |r| > 1 it keeps growing, so the sum grows without bound
The Sum Formula for a Geometric Series
Partial sum
Sn = a(1 - rn)1 - r (r ≠ 1)
The sum of the first n terms of a geometric series with first term a and ratio r
Sum of an infinite geometric series
a1 - r (|r| < 1)
As n → ∞, rⁿ → 0, so the partial sums converge to a/(1 − r)
💡 Where the formula comes from
①In Sₙ = a(1 − rⁿ)/(1 − r), if |r| < 1, then as n → ∞, rⁿ → 0
②So S = a/(1 − r)
③Since rⁿ does not go to 0 when |r| ≥ 1, the formula cannot be used there
Testing for Convergence
Test using the limit of the terms
limn→∞ an ≠ 0 ⇒ Σan diverges
If the terms do not converge to 0, the series must diverge. The converse fails: terms going to 0 do not guarantee that the series converges
Comparison test
0 ≤ an ≤ bn : Σbn converges ⇒ Σan converges
If a series with larger terms converges, the series with smaller terms converges too. Equivalently, if the smaller one diverges, so does the larger one
⚖️ Things to keep in mind
①If the terms do not go to 0, at least a fixed amount piles up again and again, so the sum grows without bound
②Some series, such as the one with terms 1/n, diverge even though their terms go to 0, so terms tending to 0 is not enough for convergence
③A series with nonnegative terms can never exceed the sum of a series whose terms are each larger, so if the larger one converges, the smaller one does too
Worked Examples
Example 1
Find the sum of the infinite geometric series Σn=1∞ (2/3)n.
1
Identify the first term a and the ratio r.
a = 23, r = 23 (|r| < 1)
2
Since |r| < 1, substitute into the sum formula S = a/(1 − r).
S = 2/31 - 2/3 = 2/31/3 = 2
▸ 2
Once the first term and the ratio are right, a geometric series goes straight into the sum formula. Because Σ starts at n = 1, the first term is (2/3)¹.
Example 2
Determine whether Σn=1∞n2n+1 converges or diverges.
1
Before trying to find a sum, examine the limit of the general term.
limn→∞n2n+1 = 12
2
The limit of the general term is not 0, so the series diverges.
lim an = 12 ≠ 0 ⇒ diverges
▸ Diverges
Before computing a sum, check whether the terms go to 0. If they do not, the series diverges at once.
Summary
When a geometric series converges
Σk=0∞ ark = a1-r (|r| < 1)
A geometric series with first term a and ratio r converges to a/(1 − r) when |r| < 1
CSAT-style problem, adapted
A geometric series with first term a and ratio r satisfies Σn=1∞ arn-1 = 9 and Σn=1∞ a2 r2(n-1) = 812. Find a.
①4
②5
③6
④7
⑤8
▸ ③ 6
1
Write the two sums as equations. The second series is geometric with first term a² and ratio r².
a1-r = 9, a21-r2 = 812
2
Dividing the second equation by the square of the first gives (1 − r)/(1 + r) = 1/2, which fixes r and then a.
1-r1+r = 12 ⇒ r = 13, a = 6
🎯 Exam Points
①A geometric series (a ≠ 0) converges if |r| < 1 and diverges if |r| ≥ 1
②The sum of an infinite geometric series is a/(1 − r)
③If lim aₙ ≠ 0, the series diverges
④Convergence of a series is defined through the limit of its partial sums Sₙ
⑤Some series diverge even though their terms go to 0