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Limit of a Geometric Sequence

Limit of a Geometric Sequence

A geometric sequence is built by multiplying by the same number r over and over. If that ratio is smaller than 1 in size, each term shrinks and eventually vanishes to 0; if it is larger than 1, the terms snowball and blow up. So the entire fate of rⁿ is decided by one thing, the size of the common ratio r, with the borders sitting exactly at r = 1 and r = -1. Slide the common-ratio r and watch whether the dots fold down toward 0, explode upward, or bounce between 1 and -1 in a diverging oscillation.

The Fate of rⁿ Is Set by r
0.7
👀 See It
①If |r|<1, each multiplication shrinks toward 0
②If r=1, it stays 1
③If r=-1, it oscillates between 1 and -1 (diverges)
④If |r|>1, it blows up and diverges
Four Cases

Classifying lim rⁿ

Chartlim rⁿ by common ratio r
Ratio rlim rⁿBehavior
|r|<10Converges
r=11Converges
r=−1NoneOscillates·diverges
|r|>1NoneDiverges
🧭 The Borders Are r=1 and r=-1
①The split happens at |r|=1
②Convergence holds only on the single band −1<r≤1
③r=−1 oscillates, so it is not convergence
④This boundary sense is the key to every geometric-limit problem
The Convergence Condition in One Line
Convergence condition for rⁿ
limn→∞ rn converges ⇔ -1 < r ≤ 1
The limit exists only on this range (0 if −1<r<1, 1 if r=1)
Key limit value
-1<r<1 ⇒ limn→∞ rn = 0
A ratio with absolute value below 1 vanishes to 0 under repeated powers
Rational Limits — Divide by the Dominant Term
✏️ Rational Limits — Divide by the Dominant Term
When rⁿ appears in numerator and denominator, divide by the term with the largest |r| to create parts that go to 0.
Example 1
For r>1, find limn→∞ rn - 1rn + 1.
1
Divide numerator and denominator by the dominant term rⁿ.
rn - 1rn + 1 = 1 - (1/r)n1 + (1/r)n
2
Since r>1 gives 1/r<1, we have (1/r)ⁿ→0.
1 - 01 + 0 = 1
1
Deciding "which term grows faster" and dividing by it is the standard move for rational limits.
Wrap-up
Geometric-sequence limit summary
limn→∞ rn = 0 (−1<r<1), 1 (r=1), diverges (otherwise)
Convergence on −1<r≤1; within it only r=1 gives 1, the rest give 0
2022 CSAT Math (Calculus) #23, adapted
Find limn→∞ 3n+1 + 2n3n - 2n.
0
1
2
3
Diverges
④ 3
1
Divide numerator and denominator by the largest base 3ⁿ.
3·3n + 2n3n - 2n = 3 + (2/3)n1 - (2/3)n
2
Since (2/3)ⁿ→0, the limit is 3/1.
3 + 01 - 0 = 3
🎯 Exam Points
①lim rⁿ converges only on −1<r≤1
②For rational forms, divide by the largest base
③Engineer (small base/large base)ⁿ→0
④Do not forget r=−1 oscillation is divergence
⑤If n sits in an exponent, suspect a geometric-sequence limit first
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Limit of a Sequence
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Series
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