Why the Common Ratio Determines the Limit
A geometric sequence is built by multiplying each term by the same number. That repeated multiplication makes its long-term behavior especially clear: if the multiplier shrinks magnitudes, the terms can approach zero; if it preserves or enlarges magnitudes, other outcomes occur. A negative multiplier adds alternating signs, but alternating signs alone do not tell us whether a sequence converges.
The question here concerns the limit of the terms of a geometric sequence. It does not require adding the terms. A geometric sequence and a geometric series are related ideas, but the convergence of a sequence is a separate question from the convergence of a series.
Definition and Explicit Formula
Repeatedly applying the common-ratio rule gives \(a_2=a_1r\), \(a_3=a_1r^2\), and, in general, \(a_n=a_1r^{n-1}\). The exponent is \(n-1\) because the first term has been multiplied by \(r\) zero times.
This formula also works when \(r=0\): it gives \(a_1\) at \(n=1\), and \(0\) for every \(n\ge 2\). The recurrence and explicit formula describe the same sequence, so either can be used to study its behavior.
Classifying Geometric Sequence Limits
For a sequence to converge to a finite real number \(L\), its terms must eventually become as close to \(L\) as desired. Precisely, for every \(\varepsilon>0\), there must be an integer \(N\) such that \(n\ge N\) implies \(|a_n-L|<\varepsilon\). The common ratio determines whether this condition can hold.
- When \(|r|<1\), the sequence converges to \(0\).
- When \(r=1\), the sequence is constant and converges to \(a_1\).
- When \(r=-1\), the terms alternate between \(a_1\) and \(-a_1\), so the sequence does not converge.
- When \(|r|>1\), the magnitudes of the terms grow without bound, so the sequence does not converge to a finite real number. For \(r<-1\), the signs alternate as the magnitudes grow.
The absolute value \(|r|\) measures how the magnitude changes at each step. When \(0<|r|<1\), each multiplication reduces the magnitude by a fixed factor. The sign of \(r\) determines whether the signs stay the same or alternate, but it does not change the fact that the magnitudes shrink.
Why the Classification Works
The key calculation is that the magnitude of the \(n\)th term is \(|a_1||r|^{n-1}\). If \(|r|<1\), the powers of \(|r|\) approach zero. If \(|r|>1\), those powers grow without bound. The boundary values \(r=1\) and \(r=-1\) preserve the magnitude rather than shrinking it.
From the explicit formula, \(|a_n|=|a_1||r|^{n-1}\).
Let \(q=|r|\) with \(0<q<1\). Then \(1/q=1+h\) for some \(h>0\), so \(q^{n-1}=1/(1+h)^{n-1}\).
Bernoulli's inequality gives \((1+h)^{n-1}\ge 1+(n-1)h\). Therefore \(0\le q^{n-1}\le 1/[1+(n-1)h]\), and the right side approaches zero.
Multiplying the bound by \(|a_1|\) shows \(|a_n|\to 0\). Since \(|a_n-0|=|a_n|\), this is exactly convergence of \(a_n\) to zero. If \(r=0\), the terms are zero from \(n=2\) onward, so the conclusion holds directly.
If \(r=1\), every term equals \(a_1\); if \(r=-1\), odd and even terms are \(a_1\) and \(-a_1\). If \(|r|>1\), put \(q=|r|=1+h\), where \(h>0\). Then \(q^{n-1}\ge 1+(n-1)h\), so \(|a_n|\) grows without bound when \(a_1\ne 0\).
For \(r=-1\), the odd-indexed terms are all \(a_1\), while the even-indexed terms are all \(-a_1\). When \(a_1\ne 0\), these are different values, so the sequence cannot approach one limit. For \(|r|>1\), the unbounded magnitudes also rule out a finite limit. If \(r<-1\), the terms additionally switch signs at every step.
Worked Examples
Worked Example: A Positive Ratio with Magnitude Less Than One
Find the limit of \(a_n=18(1/3)^{n-1}\). Here \(a_1=18\), \(r=1/3\), and \(|r|=1/3<1\), so the classification predicts a limit of \(0\).
Calculate the first terms: \(a_1=18(1/3)^0=18\); \(a_2=18(1/3)^1=6\); \(a_3=18(1/3)^2=2\); and \(a_4=18(1/3)^3=2/3\). Each term is one third of the preceding term. In general, \(|a_n|=18(1/3)^{n-1}\), which approaches zero as \(n\) increases.
Check: The proposed limit is \(0\), and the terms' magnitudes shrink to zero, so \(|a_n-0|\to 0\). The recurrence check also works: \((1/3)a_n=18(1/3)^n=a_{n+1}\). Thus the formula and limit agree.
Worked Example: A Negative Ratio with Magnitude Less Than One
Find the limit of \(b_n=8(-1/2)^{n-1}\). The ratio is negative, so signs alternate, but \(|-1/2|=1/2<1\). The terms therefore converge to \(0\).
Compute successive terms: \(b_1=8(-1/2)^0=8\); \(b_2=8(-1/2)^1=-4\); \(b_3=8(-1/2)^2=2\); \(b_4=8(-1/2)^3=-1\); and \(b_5=8(-1/2)^4=1/2\). The signs alternate, while the magnitudes are \(8,4,2,1,1/2,\ldots\) and shrink by a factor of \(1/2\).
Check: The proposed limit is \(0\), and \(|b_n-0|=8(1/2)^{n-1}\to 0\). Also, multiplying any term by the common ratio gives the next term: \((-1/2)b_n=8(-1/2)^n=b_{n+1}\).
Worked Example: A Ratio Equal to Negative One
Classify \(c_n=5(-1)^{n-1}\). Since \(r=-1\), compute the first terms: \(c_1=5\); \(c_2=-5\); \(c_3=5\); and \(c_4=-5\). More generally, \(c_{2k-1}=5\) and \(c_{2k}=-5\) for every positive integer \(k\).
The odd terms remain \(5\), and the even terms remain \(-5\). Because these two values differ, there is no single number to which all the terms approach. The sequence does not converge.
Check: If the sequence had a limit \(L\), both the odd-indexed and even-indexed terms would have to approach \(L\). But the odd terms equal \(5\) throughout and the even terms equal \(-5\) throughout. They cannot approach the same \(L\), so the nonconvergence conclusion is verified.
Worked Example: A Negative Ratio with Magnitude Greater Than One
Classify \(d_n=3(-2)^{n-1}\). Here \(|r|=2>1\), so the term magnitudes grow without bound. Calculate: \(d_1=3\); \(d_2=3(-2)=-6\); \(d_3=3(-2)^2=12\); \(d_4=3(-2)^3=-24\); and \(d_5=3(-2)^4=48\). Each step doubles the magnitude and reverses the sign.
In general, \(|d_n|=3\cdot 2^{n-1}\), which increases without bound. The signs alternate, so the sequence has no finite limit; in particular, its behavior is not convergence to zero.
Check: The common-ratio rule holds because \((-2)d_n=3(-2)^n=d_{n+1}\). The magnitude calculation gives \(3,6,12,24,48,\ldots\), confirming both the growth and the classification.
A Common Mistake: Confusing Alternation with Divergence
A negative common ratio does not automatically make a sequence divergent. For example, in the sequence \(8(-1/2)^{n-1}\), signs alternate, but the magnitudes approach zero, so the sequence converges. The relevant condition is the magnitude of the ratio: a negative \(r\) with \(|r|<1\) gives convergence to zero.
The reverse mistake is to treat \(r=-1\) as though its alternating terms cancel into a limit of zero. Pairing terms or observing that opposite values occur does not change the sequence's terms: they keep switching between \(a_1\) and \(-a_1\). A sequence converges only if its terms approach one value, not merely because different terms can be grouped into pairs.
Check Your Understanding
For each question, identify the common ratio and explain the sequence's behavior, not just the first few terms.
- Find the limit of \(a_n=10(2/5)^{n-1}\).
- Does \(b_n=6(-3/4)^{n-1}\) converge? If so, to what limit?
- Classify \(c_n=2(-1)^{n-1}\). What do its odd and even terms do?
- For \(d_n=-4(3)^{n-1}\), describe the behavior of its magnitudes and signs.
- What is the limit of \(e_n=7(0)^{n-1}\), taking the first term from the defining formula at \(n=1\)?
