Why the Ratio Test Works

Why Successive Ratios Reveal Convergence

For a series \(\sum a_n\), the size of the individual terms matters, but so does how quickly their sizes change. If each term in a tail is at most a fixed fraction of the preceding term, then the tail is controlled by a geometric series. If each term is eventually larger than its predecessor by a fixed factor greater than one, the terms cannot approach zero.

The ratio test formalizes these observations. Its proof is a useful example of a general strategy: turn information about successive terms into a bound on all later terms, then compare with a series whose behavior is known.

Definition: A series \(\sum_{n=1}^{\infty} a_n\) converges absolutely if the series of nonnegative terms \(\sum_{n=1}^{\infty}|a_n|\) converges. Absolute convergence implies convergence of the original series.

The Ratio Test

Suppose \(a_n\) is nonzero for all sufficiently large \(n\), so the ratios \(|a_{n+1}/a_n|\) are defined on a tail. If the limit

L=\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|

exists, the ratio test gives these conclusions:

If \(L<1\), then \(\sum a_n\) converges absolutely.
If \(L>1\), including \(L=+\infty\), then \(\sum a_n\) diverges.
If \(L=1\), the test gives no conclusion.

It is enough that the terms be nonzero eventually; any finite initial segment has no effect on whether an infinite series converges. If the terms are eventually zero, the series is a finite sum from that point onward and therefore converges.

Why a Ratio Below One Gives Convergence

Assume \(L<1\). There is a number \(q\) strictly between \(L\) and \(1\). Because the ratios approach \(L\), they are eventually less than \(q\). Thus, for some integer \(N\), every \(n\ge N\) satisfies \(|a_{n+1}|<q|a_n|\). Repeating this bound controls each later term by a power of \(q\) times the first term in the tail.

Choose a geometric factor.
Select \(q\) with \(L<q<1\). Since the ratios tend to \(L\), there is an \(N\) such that \(|a_{n+1}/a_n|<q\) for every \(n\ge N\).

Iterate the ratio bound.
For \(k\ge 0\), repeated multiplication gives \(|a_{N+k}|\le |a_N|q^k\). For \(k=0\), this is equality; each subsequent step uses one more ratio bound.

Compare the tail.
The geometric series with terms \(|a_N|q^k\) converges because \(0<q<1\). The comparison test therefore shows that \(\sum_{n=N}^{\infty}|a_n|\) converges.

Include the finite beginning.
Adding the finitely many terms before \(N\) does not change convergence. Hence \(\sum |a_n|\), and therefore \(\sum a_n\), converges.

The comparison in Step 3 is justified term by term: for every \(k\ge0\), \(0\le |a_{N+k}|\le |a_N|q^k\). The comparison series has sum \(|a_N|/(1-q)\), so the absolute tail is bounded by a finite number.

Key idea: A ratio limit below one does not mean the terms are exactly geometric. It means that, after some index, they decrease at least as fast as a geometric sequence with a fixed ratio less than one.

Why a Ratio Above One Gives Divergence

Now assume \(L>1\). Choose \(q\) with \(1<q<L\). Eventually every ratio is greater than \(q\), so the magnitudes of successive terms grow by at least this fixed factor. If the limit is \(+\infty\), choose any fixed \(q>1\); the same eventual bound follows from the definition of divergence to infinity.

For an index \(N\) after which the bound holds, iteration gives \(|a_{N+k}|\ge |a_N|q^k\) for every \(k\ge0\). Since \(|a_N|>0\) and \(q>1\), these lower bounds do not approach zero. Thus \(a_n\) does not tend to zero. A necessary condition for convergence of a series is that its terms tend to zero, so the series diverges.

Necessary condition: If \(\sum a_n\) converges, then \(a_n\to0\). Indeed, if \(s_n=\sum_{j=1}^n a_j\) converges to \(s\), then \(a_n=s_n-s_{n-1}\to s-s=0\). Therefore, terms that fail to tend to zero prove divergence.

Worked Examples

Worked Example: A Ratio Limit Below One

Determine whether \(\sum_{n=1}^{\infty} n/2^n\) converges.

Let \(a_n=n/2^n\), which is positive for \(n\ge1\). Its successive ratio is

\frac{a_{n+1}}{a_n} = \frac{(n+1)/2^{n+1}}{n/2^n} = \frac{n+1}{2n} = \frac12+\frac{1}{2n}.

As \(n\to\infty\), \(1/(2n)\to0\), so the ratio limit is \(1/2<1\). The ratio test already proves absolute convergence. To see the geometric bound directly, for every \(n\ge2\),

\frac{a_{n+1}}{a_n} = \frac12+\frac{1}{2n} \le \frac12+\frac14 =\frac34.

Also \(a_2=2/4=1/2\). Repeated application of the bound gives \(a_{2+k}\le (1/2)(3/4)^k\) for every \(k\ge0\). Therefore,

\sum_{n=1}^{\infty}a_n \le a_1+\frac12\sum_{k=0}^{\infty}\left(\frac34\right)^k = \frac12+\frac12\cdot\frac{1}{1-3/4} = \frac12+2 = \frac52.

This confirms convergence by an explicit finite bound on the sum. The ratio bound and the geometric-series formula verify the comparison used.

Worked Example: A Ratio Limit Greater Than One

Determine whether \(\sum_{n=1}^{\infty} n!/5^n\) converges.

Set \(a_n=n!/5^n\). Then

\frac{a_{n+1}}{a_n} = \frac{(n+1)!/5^{n+1}}{n!/5^n} = \frac{(n+1)n!}{5\cdot 5^n}\cdot\frac{5^n}{n!} = \frac{n+1}{5}.

The ratio \((n+1)/5\) tends to \(+\infty\). More explicitly, for every \(n\ge5\), \((n+1)/5\ge6/5>1\). The fifth term is

a_5=\frac{5!}{5^5}=\frac{120}{3125}=\frac{24}{625}>0.

Iterating the ratio bound gives \(a_{5+k}\ge(24/625)(6/5)^k\) for every \(k\ge0\). In particular, these terms do not tend to zero. The necessary condition for series convergence fails, so \(\sum n!/5^n\) diverges. Substituting \(k=0\) into the bound gives \(a_5\ge24/625\), as required, and each later step multiplies by at least \(6/5\).

Worked Example: A Ratio Limit of One

A ratio limit of one is inconclusive. Compare the two positive-term series \(\sum 1/n\) and \(\sum 1/n^2\).

For \(a_n=1/n\),

\frac{a_{n+1}}{a_n} = \frac{1/(n+1)}{1/n} = \frac{n}{n+1} \longrightarrow 1.

This series diverges. For example, grouping terms in blocks between successive powers of two gives, for each integer \(j\ge0\),

\sum_{n=2^j+1}^{2^{j+1}}\frac1n \ge 2^j\cdot\frac{1}{2^{j+1}} = \frac12.

There are infinitely many such blocks, each contributing at least \(1/2\), so the partial sums are unbounded.

For \(b_n=1/n^2\),

\frac{b_{n+1}}{b_n} = \frac{1/(n+1)^2}{1/n^2} = \left(\frac{n}{n+1}\right)^2 \longrightarrow 1.

Yet this series converges: for \(n\ge2\), \(1/n^2\le1/[n(n-1)]\), and

\sum_{n=2}^{M}\frac{1}{n(n-1)} = \sum_{n=2}^{M}\left(\frac{1}{n-1}-\frac1n\right) = 1-\frac1M < 1.

The partial sums of \(\sum 1/n^2\) are bounded above by \(1+1=2\), and they are increasing, so the series converges. Thus both ratio limits equal one, although one series diverges and the other converges. The ratio test cannot decide between them.

A Common Misinterpretation

A frequent mistake is to treat \(L=1\) as a borderline version of the convergence case and conclude that the series converges. For the harmonic series, the ratio tends to one, but the series diverges, as the block calculation above shows. The opposite incorrect conclusion is also possible: inferring divergence whenever \(L=1\), even though \(\sum 1/n^2\) converges.

Correction: The ratio test proves absolute convergence only when the ratio limit is strictly less than one, and divergence only when it is strictly greater than one. At exactly one, use another argument or test.

What the Proof Establishes

The ratio test succeeds because a limit gives a uniform bound on every sufficiently late ratio. When that bound is below one, multiplying it repeatedly produces a geometric upper bound for the terms, and the geometric tail has finite sum. When the bound is above one, repeated multiplication prevents the terms from tending to zero. These arguments explain both the conclusions and the strict inequalities in the test.

The test is especially useful when factorials, exponentials, or products make the quotient of consecutive terms simpler than the terms themselves. Its result should still be interpreted carefully: a limit of one is not a verdict about the series, but a signal that this test has reached its limit.

Check Your Understanding

Use the ratio-test proof and its conditions to answer these questions.

Why can a number \(q\) be chosen with \(L<q<1\) when \(L<1\), and what does convergence of the geometric comparison establish?
If the ratio limit is greater than one, why does iteration show that the terms fail the necessary condition for series convergence?
What assumption about the terms is needed in order to form the successive ratios on a tail?
For \(a_n=4^n/n!\), compute \(|a_{n+1}/a_n|\) and determine the ratio-test conclusion.
Give one convergent series and one divergent series whose successive absolute-term ratios both tend to one.

What You'll Learn