When a Limit Greater Than One Is Decisive
A ratio or root limit greater than one has a different role from the familiar case where the limit is less than one. Instead of proving convergence, it shows that the terms of the series cannot approach zero. That is enough to prove divergence.
The key point is that convergence of a series requires its individual terms to tend to zero. A ratio or root limit greater than one prevents that necessary condition from holding. By contrast, a limit equal to one does not settle the question: series with that same limit can converge or diverge.
The Necessary Condition for a Convergent Series
For a series \(\sum_{n=1}^{\infty} a_n\), let \(S_N=\sum_{n=1}^{N}a_n\) denote its \(N\)th partial sum. If the series converges, then \(S_N\) approaches a finite limit \(S\). For \(N\ge2\), the terms satisfy \(a_N=S_N-S_{N-1}\). Both partial sums on the right approach \(S\), so their difference approaches zero.
This is a necessary condition, not a sufficient one: terms tending to zero do not by themselves guarantee convergence. But if the terms do not tend to zero, the series must diverge. The ratio and root arguments in this tutorial work by establishing precisely that failure.
How a Ratio Limit Greater Than One Forces Divergence
Suppose the ratio limit is greater than one. Choose a fixed number \(q\) strictly between one and that limit. Eventually, the ratios of successive term magnitudes exceed \(q\). Thus each term’s magnitude is at least \(q\) times the preceding magnitude. Repeating that inequality gives a geometric lower bound, which grows rather than shrinks.
Here is the complete argument. Since \(L>1\), choose \(q\) with \(1<q<L\) when \(L\) is finite; if \(L=+\infty\), choose any fixed \(q>1\). By the definition of the limit, there is an index \(N\) such that for every \(n\ge N\), \(|a_{n+1}/a_n|\ge q\). Because the ratios are defined and \(a_N\ne0\), repeated multiplication gives, for each integer \(k\ge0\),
For \(k=0\), the product is empty and equals \(1\), so equality holds. Since \(|a_N|>0\) and \(q>1\), the lower bound \(|a_N|q^k\) grows without bound as \(k\) increases. In particular, \(|a_n|\) cannot approach zero. The necessary condition for convergence therefore fails, proving divergence.
Worked Example: A Ratio Limit of Three Halves
Worked Example: A Ratio Limit of Three Halves
Consider the series with terms \(a_n=\frac{(3/2)^n}{n+1}\), for \(n\ge1\). The terms are positive, so their magnitudes are the same terms. Compute the successive-term ratio:
Taking the limit gives \(\lim_{n\to\infty}a_{n+1}/a_n=3/2>1\). To see the geometric growth explicitly, use \(q=5/4\). For \(n\ge4\),
At \(n=4\), equality holds. Also, \(a_4=(3/2)^4/5=81/80\). Therefore, for every \(n\ge4\), multiplying the ratios from index \(4\) through index \(n-1\) gives
For \(n=4\), the product has no factors and the bound is equality. For larger \(n\), the bound grows without limit. Thus \(a_n\) does not approach zero, and the series diverges by the necessary condition. This conclusion agrees with the ratio-limit theorem.
How a Root Limit Greater Than One Forces Divergence
The root argument is direct. If \(\sqrt[n]{|a_n|}\) eventually exceeds a fixed number \(q>1\), raising both sides to the positive integer power \(n\) shows that \(|a_n|\) eventually exceeds \(q^n\). These magnitudes grow, so they cannot tend to zero.
To justify the theorem, choose \(q\) with \(1<q<L\) if \(L\) is finite, or any \(q>1\) if \(L=+\infty\). From the definition of the root limit, there is an \(N\) such that \(\sqrt[n]{|a_n|}>q\) for every \(n\ge N\). Raising this inequality to the \(n\)th power preserves its direction because both sides are nonnegative and the power is positive. Hence \(|a_n|>q^n\) for \(n\ge N\), so the terms cannot tend to zero. The necessary condition for convergence proves divergence.
Worked Example: A Root Limit of Two
Worked Example: A Root Limit of Two
Let \(a_n=(2+1/n)^n\), for \(n\ge1\). Its \(n\)th root is exactly
Therefore, \(\lim_{n\to\infty}\sqrt[n]{|a_n|}=2>1\). The root-limit theorem proves that the series \(\sum a_n\) diverges. In this example, the failure of the term condition can also be seen directly: \(2+1/n>2\), so \(a_n=(2+1/n)^n>2^n\ge2\) for every \(n\ge1\). The terms are therefore bounded below by \(2\), and cannot approach zero. This direct check confirms the conclusion from the root limit.
Why a Limit of One Is Inconclusive
A limit equal to one is not a weakened version of the divergence result. It simply does not provide the geometric growth needed by the proof. Ratios approaching one do not eventually stay above a fixed \(q>1\); likewise, roots approaching one do not eventually imply a lower bound \(q^n\) for a fixed \(q>1\).
More importantly, a limit of one does not determine whether a series converges. The following two series have ratio limit one and root limit one, but one converges while the other diverges.
Worked Example: Two Series with Limit One
First take \(b_n=1/[n(n+1)]\). Its successive-term ratio is
Its root limit is also one, since \(\sqrt[n]{b_n}=[n(n+1)]^{-1/n}\to1\). Yet the terms telescope: \(b_n=1/n-1/(n+1)\). Thus the partial sum is
So this series converges, as verified by its partial sums.
Now take \(c_n=1/n\). Its ratio and root limits are
But the harmonic series \(\sum c_n\) diverges. To verify this, group terms up to \(2^m\). After the first term, the terms with indices \(2^{k-1}+1\) through \(2^k\) form a group of \(2^{k-1}\) terms, each at least \(1/2^k\). Each group therefore has sum at least \(1/2\), so
These partial sums grow without bound as \(m\to\infty\). Thus the series diverges, despite both limits being one. The two checked outcomes show why a limit of one is inconclusive.
A Common Mistake: Treating One as a Verdict
A common error is to calculate a ratio limit of one and announce divergence, perhaps reasoning that the ratios are “not less than one.” For example, the terms \(b_n=1/[n(n+1)]\) above have ratio limit one, but their partial sums approach one. The claim of divergence based on that limit is wrong.
The correction is to use the tests only for the conclusions their hypotheses support. A ratio or root limit strictly greater than one proves that the terms fail to approach zero and hence proves divergence. A limit strictly less than one is the case used to prove absolute convergence. A limit equal to one gives neither conclusion; another argument is needed. No conclusion about the series follows from the value one alone.
Check Your Understanding
For each question, identify exactly what the stated limit does and does not establish.
- If \(\lim |a_{n+1}/a_n|=7/5\), what geometric lower bound can be obtained after choosing a suitable \(q\), and what does it imply for \(\sum a_n\)?
- Why must the ratio \(a_{n+1}/a_n\) be defined for all sufficiently large \(n\) to apply the ratio-limit divergence argument?
- If \(\lim\sqrt[n]{|a_n|}=+\infty\), explain how to prove that \(a_n\) does not tend to zero.
- Can a series whose ratio limit is one converge? Give the example from this tutorial and state how its convergence was checked.
- Can a series whose root limit is one diverge? Give the example from this tutorial and state why its partial sums do not remain bounded.
