Proving a Geometric Sequence Converges

How the Common Ratio Controls Convergence

An arithmetic sequence changes by a fixed amount at each step. A geometric sequence instead multiplies by a fixed ratio, so its terms may shrink toward zero, alternate in sign, or grow in magnitude. The explicit formula established in the earlier tutorial on geometric sequences lets us distinguish these possibilities: if the initial value is \(a_0\) and the common ratio is \(r\), then \(a_n=a_0r^n\).

The key quantity is \(|r|\), but the cases \(|r|=1\) require care. A ratio of \(-1\) makes nonzero terms alternate rather than settle. A ratio of \(1\) gives a constant sequence. If the initial value is zero, however, every term is zero regardless of the ratio. We will prove the decay case directly from the epsilon-N definition, then use boundedness and subsequences to rule out the other cases that do not converge.

We first establish the estimate behind the main convergence argument. When \(0<q<1\), the powers \(q^n\) become small because their reciprocals grow at least linearly. The elementary inequality used to show this is included here so that the threshold in the convergence proof is justified.

A Decay Estimate for Powers

Proof. We use induction. At \(n=0\), both sides equal \(1\). Suppose \((1+h)^n\geq 1+nh\). Since \(1+h>0\), multiplication preserves the inequality, and

The final inequality holds because \(n\geq0\) and \(h^2>0\). This proves the claim for \(n+1\), so induction establishes the inequality for every \(n\in\mathbb{N}_0\). \(\square\)

Now take \(q\) with \(0<q<1\), and set \(h=1/q-1\), which is positive. Then \(1/q=1+h\). Since \(q^n>0\), taking reciprocals of the positive quantities in the lemma gives

The right side can be made smaller than any prescribed positive tolerance by choosing \(n\) sufficiently large. This yields a complete epsilon-N proof rather than merely an informal claim that repeated multiplication makes the terms small.

Proof. First suppose \(q=0\). Then \(q^n=0\) for every \(n\geq1\). Given \(\varepsilon>0\), choose \(N=1\). For every \(n\geq N\), \(|q^n-0|=0<\varepsilon\), so \(q^n\to0\).

Now suppose \(0<q<1\), and let \(\varepsilon>0\). Define \(h=1/q-1>0\). By the Archimedean property, choose \(N\in\mathbb{N}_0\) such that \(N>(1/\varepsilon-1)/h\). If the quantity on the right is negative, \(N=0\) already satisfies this strict inequality; otherwise the Archimedean property supplies such a nonnegative integer. For every \(n\geq N\), we have \(nh\geq Nh\), and the choice of \(N\) implies \(1+Nh>1/\varepsilon\). The estimate above therefore gives

Thus the epsilon-N condition holds for every positive \(\varepsilon\), and \(q^n\to0\). \(\square\)

When the Ratio Has Absolute Value Less Than One

If \(|r|<1\), the absolute value of \(a_0r^n\) is \(|a_0||r|^n\). When \(a_0=0\), every term is zero. When \(a_0\neq0\), the theorem on powers below one shows that \(|r|^n\to0\); we can also choose a threshold explicitly by applying the same estimate with tolerance \(\varepsilon/|a_0|\). The absolute value is useful because it handles positive and negative ratios in one argument.

Proof. Let \(\varepsilon>0\). If \(a_0=0\), then \(a_n=0\) for every \(n\), and \(N=0\) gives \(|a_n-0|=0<\varepsilon\) for every \(n\geq N\).

Suppose instead that \(a_0\neq0\). Put \(q=|r|\), so \(0\leq q<1\). If \(q=0\), then \(r=0\), and \(a_n=a_0r^n=0\) for every \(n\geq1\). Taking \(N=1\) proves convergence to zero. If \(0<q<1\), apply the powers-below-one theorem with the positive tolerance \(\varepsilon/|a_0|\). It gives an \(N\in\mathbb{N}_0\) such that \(q^n<\varepsilon/|a_0|\) whenever \(n\geq N\). Since \(|r^n|=|r|^n=q^n\), for every such \(n\),

In every case the epsilon-N condition holds, so \(a_n\to0\). \(\square\)

Ratios on or Outside the Unit Circle

The cases \(|r|\geq1\) do not all have the same outcome. If \(r=1\), then \(a_n=a_0\) for every \(n\), so the constant-sequence theorem applies. If \(a_0=0\), the sequence is again identically zero for every value of \(r\). For a nonzero initial value, the ratio \(r=-1\) produces two alternating values and cannot converge.

When \(|r|>1\), the magnitudes grow without bound. To verify this without assuming a later theorem about exponential growth, put \(s=|r|>1\) and \(h=s-1>0\). The linear lower bound for powers gives \(s^n=(1+h)^n\geq1+nh\). If \(a_0\neq0\), it follows that \(|a_n|=|a_0|s^n\) is unbounded. Since every convergent sequence is bounded, such a geometric sequence cannot converge.

Proof. First consider the listed cases. If \(a_0=0\), then \(a_n=0\) for every \(n\), so the sequence converges to zero. If \(|r|<1\), the theorem on contracting ratios proves convergence to zero. If \(r=1\), then \(a_n=a_0\) for every \(n\), so the constant-sequence theorem gives convergence to \(a_0\).

For the converse, suppose none of these conditions holds. Then \(a_0\neq0\), \(|r|\geq1\), and \(r\neq1\). If \(r=-1\), the even and odd subsequences are \(a_{2k}=a_0(-1)^{2k}=a_0\) and \(a_{2k+1}=a_0(-1)^{2k+1}=-a_0\). If the original sequence converged to \(L\), the earlier theorem on subsequences of a convergent sequence would imply that both subsequences converge to \(L\). But constant sequences have their constant values as their limits, so uniqueness of limits would give \(L=a_0\) and \(L=-a_0\). These equalities imply \(a_0=0\), contrary to the assumption. Therefore the sequence does not converge when \(r=-1\).

The remaining possibility is \(|r|>1\). Set \(s=|r|=1+h\), where \(h>0\). For every \(n\in\mathbb{N}_0\),

Because \(|a_0|h>0\), the Archimedean property gives indices \(n\) for which the right side exceeds any prescribed real bound. Thus \((a_n)\) is unbounded. Every convergent sequence is bounded, so \((a_n)\) cannot converge. This exhausts all cases not listed in the criterion and proves the theorem. \(\square\)

Using the Criterion Carefully

The common ratio alone does not always decide convergence: the initial value matters when the ratio has magnitude at least one. For example, a ratio of \(4\) gives a divergent sequence from any nonzero initial value, but an initial value of zero produces the constant zero sequence. It is therefore safest to check \(a_0=0\) before classifying the ratio.

The boundary cases also deserve separate attention. A ratio of \(1\) gives a constant sequence, which converges to its initial value. A ratio of \(-1\) does not converge when the initial value is nonzero, because its even and odd terms remain at two distinct values. A ratio with magnitude greater than one is ruled out by unboundedness, whereas a ratio with magnitude less than one gives convergence to zero, including when the ratio is negative.

A common pitfall is to treat alternating signs as evidence of nonconvergence by themselves. The sequence \(4(-1/3)^n\) alternates, yet it converges to zero because the magnitudes shrink. Another is to claim that a ratio of magnitude at least one always forces divergence; this overlooks the zero sequence when \(a_0=0\), and the constant sequence when \(r=1\). The complete criterion accounts for each exception explicitly.