Adding infinitely many numbers seems as if it must give something infinitely large, but it does not have to.
If the ratio is less than 1 in absolute value, each new term is smaller than the one before by a fixed factor, so however many you add, the total never passes a certain value and keeps moving toward it.
That value is a/(1 − r), the first term a divided by 1 − r.
The picture uses a = 1 and r = 0.6: as you increase the number of terms n, the partial-sum bars keep growing but never get past the red line at 2.5.
Adding Forever, Yet There Is an End
seegongsik.com
With a = 1 and r = 0.6, the partial sums Sₙ approach the limit a/(1 − r) = 2.5
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👀 What to read in the picture
①Bar Sₙ is the sum of the first n terms, and the bright top piece is the newly added term 0.6ⁿ⁻¹
②Each new piece is 0.6 times the one before, so the bars grow by less and less
③The remaining gap 2.5 − Sₙ in the top band equals 2.5 × 0.6ⁿ, which approaches 0 as n grows. That is why the bars never pass the red line at 2.5 and converge to it
Start from the Partial Sum
Partial sum of a geometric sequence
Sn = a(1 - rn)1 - r (r ≠ 1)
The sum of the first n terms of a geometric sequence with first term a and ratio r
🧱 First, the sum of n terms
①The sum of an infinite series is defined as the limit of its partial sums Sₙ
②First write Sₙ with the sum formula for a geometric sequence
③Then take the limit as n → ∞. Keeping this order stops you from assigning a sum to a series that does not converge
Take the Limit and the Formula Appears
Sum of an infinite geometric series
S = a1 - r (|r| < 1)
With first term a ≠ 0 and ratio r, the series converges only when |r| < 1, and its sum is a/(1 − r)
Condition for convergence
converges ⇔ a = 0 or |r| < 1
If the first term is 0, every term is 0 and the sum is 0 whatever the ratio. If a ≠ 0, you need |r| < 1
💨 rⁿ disappears
①If |r| < 1, then as n → ∞ we get rⁿ → 0
②In the partial sum a(1 − rⁿ)/(1 − r), rⁿ goes to 0, leaving only a/(1 − r)
③If |r| ≥ 1, rⁿ does not go to 0, so as long as a ≠ 0 the series diverges
Try It
Example 1
Find the sum of the infinite geometric series with first term 3 and ratio 1/3.
1
Check the size of the ratio. Since |1/3| < 1, the series converges.
2
Substitute a = 3 and r = 1/3 into S = a/(1 − r).
S = 31 - 1/3 = 32/3
▸ S = 9/2
Checking |r| < 1 before using the formula prevents mistakes.
Summary
Key result
|r|<1 ⇒ ∑n=1∞ a rn-1 = a1 - r
An infinite geometric series with first term a and ratio r converges to a/(1 − r) when |r| < 1
2023 KICE mock exam Math (Calculus) type, adapted
Find ∑n=1∞2n3n.
①1
②3/2
③2
④5/2
⑤Diverges
▸ ③ 2
1
Writing the general term as (2/3)ⁿ shows a geometric series with first term 2/3 and ratio 2/3.
2
Since |2/3| < 1, S = (2/3)/(1 − 2/3).
S = 2/31/3 = 2
🎯 Exam Points
①Check |r| < 1 before using the sum formula
②The sum formula is S = a/(1 − r); identify the first term and the ratio exactly
③∑arⁿ has first term ar, while ∑arⁿ⁻¹ has first term a
④Repeating decimals and total areas of shapes also come down to this formula