high school 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 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

Common ratio r0.7

👀 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

⑤(smaller/larger)ⁿ → 0. Example: 2ⁿ/3ⁿ → 0. The sketch is a single rⁿ

Four Cases

Classifying lim rⁿ

lim rⁿ by common ratio r

Ratio r	lim rⁿ	Behavior
\|r\|<1	0	Converges
r=1	1	Converges
r=−1	None	Oscillates·diverges
\|r\|>1	None	Diverges

🧭 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

lim_n→∞ 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 ⇒ lim_n→∞ 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 lim_n→∞ rⁿ - 1rⁿ + 1.

Divide numerator and denominator by the dominant term rⁿ.

rⁿ - 1rⁿ + 1 = 1 - (1/r)ⁿ1 + (1/r)ⁿ

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

lim_n→∞ 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

exam-style

Find lim_n→∞ 3ⁿ⁺¹ + 2ⁿ3ⁿ - 2ⁿ.

①0

②1

③2

④3

⑤Diverges

▸ ④ 3

Divide numerator and denominator by the largest base 3ⁿ.

3·3ⁿ + 2ⁿ3ⁿ - 2ⁿ = 3 + (2/3)ⁿ1 - (2/3)ⁿ

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

← Previous

Limit of a Sequence

Series

Was this helpful? Support seegongsik