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Ratio, Root, and Test Selection · Tutorial 691 of 998

Finding Parameter Thresholds with the Ratio Test

Use the ratio test to find convergence thresholds in parameters, then analyze the cases where its ratio limit is inconclusive.

Advanced 11 min read

What You'll Learn

  • Compute the ratio of successive absolute terms when powers of factorials are present.
  • Determine how the sign of the factorial exponent controls the ratio limit.
  • Find the threshold for convergence when the factorial exponent is zero.
  • Analyze the boundary cases where the ratio test gives a limit of one.
  • Distinguish absolute convergence, conditional convergence, and divergence.

Introduction

When a series contains parameters, convergence can depend sharply on their values. A factorial in the denominator can force rapid decay, while an exponential factor can either help or oppose that decay. The ratio test is especially useful for comparing these effects and locating parameter thresholds.

A ratio limit of one, however, does not settle convergence. It marks a boundary that must be checked separately. For series with terms involving \(n^p c^n/(n!)^q\), this distinction is central: the ratio test settles most parameter ranges, while the remaining cases reduce to polynomial or alternating-series behavior.

The family of series and its ratio

Let \(p,q,c\in\mathbb R\), and consider the series

$$ \sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty}\frac{n^p c^n}{(n!)^q}. $$

Since \(n!\) is positive for every \(n\in\mathbb N\), its real power \((n!)^q\) is defined and positive. The magnitude of the \(n\)th term is

$$ u_n=|a_n|=\frac{n^p |c|^n}{(n!)^q}. $$

When \(c\ne0\), each \(u_n\) is positive, so the ratio \(u_{n+1}/u_n\) is defined. Using \((n+1)!=(n+1)n!\), we obtain

$$ \frac{u_{n+1}}{u_n} = \frac{(n+1)^p|c|^{n+1}}{((n+1)!)^q} \frac{(n!)^q}{n^p|c|^n} = |c|\left(\frac{n+1}{n}\right)^p\frac{1}{(n+1)^q} = |c|\frac{(1+1/n)^p}{(n+1)^q}. $$
Ratio test: For a series with nonzero terms, if \(\lim_{n\to\infty}|a_{n+1}/a_n|=L<1\), the series converges absolutely. If the limit is greater than \(1\), or is infinite, the terms do not tend to zero and the series diverges. If the limit is \(1\), the test gives no conclusion.

The factor \((1+1/n)^p\) tends to \(1\) for every fixed real \(p\). Thus, when \(c\ne0\), the ratio is governed by the remaining factor \(|c|/(n+1)^q\): it tends to zero if \(q>0\), equals \(|c|\) in the limit if \(q=0\), and tends to infinity if \(q<0\).

Why the factorial exponent creates the threshold

The ratio calculation explains the main division of cases. If \(q>0\), each step introduces a factor \((n+1)^{-q}\), which tends to zero. A fixed exponential factor \(|c|\) and the polynomial factor \((1+1/n)^p\), which tends to \(1\), cannot prevent the ratio from tending to zero. If \(q<0\), the factor \((n+1)^{-q}\) instead grows without bound. The terms then eventually increase in magnitude rather than approaching zero.

Threshold principle: For \(c\ne0\), positive \(q\) gives absolute convergence for every real \(p\) and every finite real \(c\). Negative \(q\) gives divergence. At \(q=0\), the outcome depends on \(c\), and when \(|c|=1\), it also depends on \(p\).

This is a threshold in \(q\), but the threshold value \(q=0\) cannot be decided by the ratio test alone. Its ratio limit may be less than, greater than, or equal to one according to \(|c|\). When it equals one, the polynomial factor must be examined directly.

Worked Example: Finding the threshold in the factorial exponent

Worked Example: Finding the threshold in the factorial exponent

Determine all real \(q\) for which the series

$$ \sum_{n=1}^{\infty}\frac{n^3 5^n}{(n!)^q} $$

converges.

Here \(p=3\) and \(c=5\ne0\). The ratio of successive absolute terms is

$$ \frac{u_{n+1}}{u_n} = 5\frac{(1+1/n)^3}{(n+1)^q}. $$

If \(q>0\), then \((1+1/n)^3\to1\) and \((n+1)^q\to\infty\), so the ratio tends to \(0\). The ratio test proves absolute convergence.

If \(q=0\), the ratio tends to \(5\), which is greater than \(1\). The ratio test proves divergence.

If \(q<0\), then \((n+1)^q\to0\), so the ratio tends to infinity. In particular, it is eventually greater than \(2\); consequently the magnitudes eventually grow by a factor greater than \(2\) at each step, and the terms do not tend to zero. The series diverges.

Therefore the series converges exactly when \(q>0\). As a check, the three ratio limits are respectively \(0\), \(5\), and infinity, matching the convergence classification in each case.

The zero-factorial-exponent boundary

When \(q=0\), the factorial contributes no decay or growth because \((n!)^0=1\). For \(c\ne0\), the ratio limit is \(|c|\). If \(|c|<1\), the ratio test gives absolute convergence. If \(|c|>1\), it gives divergence. The remaining boundary is \(|c|=1\), where real \(c\) means \(c=1\) or \(c=-1\).

For \(c=1\), the series is \(\sum n^p\). The \(p\)-series criterion states that \(\sum 1/n^s\) converges exactly when \(s>1\). Taking \(s=-p\), we see that \(\sum n^p\) converges exactly when \(p<-1\).

For \(c=-1\), the series is \(\sum (-1)^n n^p\). Its absolute-value series is \(\sum n^p\), so absolute convergence again holds exactly when \(p<-1\). If \(-1\le p<0\), the magnitudes \(n^p\) decrease to zero, so the alternating series test gives convergence, but the absolute-value series diverges. This is conditional convergence. If \(p\ge0\), the term magnitudes \(n^p\) do not tend to zero, so the series diverges.

Boundary classification for \(q=0\): If \(c=1\), convergence occurs exactly when \(p<-1\). If \(c=-1\), the series converges absolutely when \(p<-1\), conditionally when \(-1\le p<0\), and diverges when \(p\ge0\).

Worked Example: A ratio limit of one at \(c=1\)

Worked Example: A ratio limit of one at \(c=1\)

Find the values of \(p\) for which

$$ \sum_{n=1}^{\infty}n^p $$

converges, and explain why the ratio test does not decide the answer.

Here \(c=1\) and \(q=0\). The terms are positive, and their successive ratio is

$$ \frac{(n+1)^p}{n^p} = \left(\frac{n+1}{n}\right)^p = (1+1/n)^p \longrightarrow 1. $$

The ratio test is therefore inconclusive for every real \(p\). Rewrite \(n^p=1/n^{-p}\). By the \(p\)-series criterion, \(\sum 1/n^s\) converges exactly when \(s>1\); here \(s=-p\). Thus convergence holds precisely when \(-p>1\), or \(p<-1\). At \(p=-1\), the series is the harmonic series and diverges; for \(p>-1\), it also diverges by the same criterion.

For a check, take \(p=-2\). The series becomes \(\sum 1/n^2\), which converges, while its ratio limit is still \(1\). Take \(p=-1\): the ratio limit remains \(1\), but the harmonic series diverges. These checks show why a ratio limit of one cannot supply the boundary decision.

Worked Example: An alternating boundary and the full range in \(c\)

Worked Example: An alternating boundary and the full range in \(c\)

For the family

$$ \sum_{n=1}^{\infty}\frac{(-1)^n c^n}{\sqrt n}, $$

classify convergence for real \(c\).

The absolute terms are \(u_n=|c|^n/\sqrt n\). If \(c\ne0\), their ratio is

$$ \frac{u_{n+1}}{u_n} = |c|\sqrt{\frac{n}{n+1}} = \frac{|c|}{\sqrt{1+1/n}} \longrightarrow |c|. $$

If \(0<|c|<1\), the ratio test proves absolute convergence. If \(|c|>1\), the ratio limit is greater than \(1\), so the terms do not tend to zero and the series diverges.

If \(c=0\), every term is zero, so the series converges. This case is handled directly rather than by a ratio, since \(u_{n+1}/u_n=0/0\) is undefined.

At \(c=1\), the series is \(\sum (-1)^n/\sqrt n\). Its term magnitudes \(1/\sqrt n\) decrease to zero, so the alternating series test proves convergence. Its absolute-value series is \(\sum 1/\sqrt n\), a \(p\)-series with exponent \(1/2\), and diverges. Thus convergence is conditional.

At \(c=-1\), the factor \((-1)^n c^n\) equals \(1\) for every \(n\), so the series is \(\sum 1/\sqrt n\), which diverges. In summary, the series converges absolutely for \(|c|<1\), converges conditionally at \(c=1\), and diverges for \(c=-1\) or \(|c|>1\). This check distinguishes the two endpoints even though both have ratio limit \(1\).

A complete classification

The preceding cases give the convergence range for the whole family. The special case \(c=0\) is stated separately because the ratio of successive absolute terms is then undefined. For all nonzero \(c\), the ratio-test calculation applies.

Parameter conditions Conclusion
\(c=0\), any real \(p,q\) Every term is zero; the series converges.
\(c\ne0,\ q>0\) The ratio limit is \(0\); the series converges absolutely.
\(c\ne0,\ q<0\) The ratio limit is infinity; the series diverges.
\(q=0,\ 0<|c|<1\) The ratio limit is less than \(1\); the series converges absolutely.
\(q=0,\ |c|>1\) The ratio limit is greater than \(1\); the series diverges.
\(q=0,\ c=1\) The series converges exactly when \(p<-1\).
\(q=0,\ c=-1\) It converges absolutely for \(p<-1\), conditionally for \(-1\le p<0\), and diverges for \(p\ge0\).

Common mistake: using the ratio test at every boundary

A common error is to calculate a ratio limit of \(1\) and then claim that the series diverges, perhaps because the terms seem not to decrease quickly enough. The ratio test makes no such claim: a limit of \(1\) is inconclusive. For example, \(\sum 1/n^2\) converges, while \(\sum 1/n\) diverges, although both have successive-term ratio tending to \(1\).

Another error is to apply the ratio formula when \(c=0\). For example, in \(\sum n^2 0^n/(n!)^q\), all terms are zero because the index begins at \(n=1\). The ratio of successive absolute terms is \(0/0\), not a usable ratio-test limit. The correct conclusion is direct: the sum is zero and therefore converges.

The practical rule is to use the ratio test to find the parameter ranges where its limit is strictly below or above one, then examine every boundary where the limit equals one. When a parameter makes the terms identically zero, handle that case directly before forming a ratio.

Check Your Understanding

Use the ratio test where it applies, and analyze any boundary cases separately.

  1. For which real \(q\) does \(\sum_{n=1}^{\infty} n^2 3^n/(n!)^q\) converge?
  2. For which real \(p\) does \(\sum_{n=1}^{\infty} n^p\) converge?
  3. Classify \(\sum_{n=1}^{\infty}(-1)^n n^{-3/4}\) as absolutely convergent, conditionally convergent, or divergent.
  4. Why is the ratio test not applicable to \(\sum_{n=1}^{\infty}n^2 0^n/(n!)^2\), and does the series converge?
  5. For \(q=0\) and \(|c|=1\), what additional information must be checked after obtaining a ratio limit of one?

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