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Ratio, Root, and Test Selection · Tutorial 679 of 998

Series with a Single Factorial in the Denominator

Learn to use successive-term ratios to classify these factorial-denominator series and understand why every fixed base gives absolute convergence when the factorial power is positive.

Advanced 9 min read

What You'll Learn

  • Compute the absolute ratio of successive terms for a factorial-denominator series.
  • Apply the ratio test when the factorial appears to a positive real power.
  • Classify the series for positive, negative, and zero values of the fixed base.
  • Prove absolute convergence by bounding the tail with a geometric series.
  • Avoid errors from mishandling the factorial or the exponent in the ratio.

Introduction

A fixed exponential factor can make the terms of a series grow quickly. A factorial in the denominator changes that behavior: each successive term gains a factor of \(n+1\) in the factorial, and a positive power of that factor eventually outweighs any fixed exponential base.

The ratio test makes this comparison precise. For the family considered here, the ratio of successive absolute terms tends to zero, regardless of the size or sign of the fixed base. We will compute that ratio, establish the full convergence classification, and explain why the conclusion follows.

The Series and Its Successive-Term Ratio

Let \(c\) be a fixed real number and let \(p>0\). Consider the series

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

Here \(n!=1\cdot2\cdots n\), and \((n!)^p\) means the positive real number obtained by raising \(n!\) to the power \(p\). Write \(a_n=c^n/(n!)^p\). When \(c\ne0\), every term is nonzero, so the ratio of successive absolute terms is defined.

$$ \frac{|a_{n+1}|}{|a_n|} = \frac{|c|^{n+1}}{((n+1)!)^p} \cdot \frac{(n!)^p}{|c|^n} = \frac{|c|}{(n+1)^p}. $$

The last equality uses \((n+1)!=(n+1)n!\). Since \(p>0\), the denominator \((n+1)^p\) grows without bound as \(n\) grows. Therefore,

$$ \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty}\frac{|c|}{(n+1)^p} =0. $$
Theorem: For every fixed real \(c\) and every fixed \(p>0\), the series \(\sum_{n=1}^{\infty}c^n/(n!)^p\) converges absolutely. If \(c=0\), all its terms are zero. If \(c\ne0\), the ratio test applies to the absolute values, and their successive-term ratio tends to \(0<1\).

Why the Ratio Test Gives Absolute Convergence

The ratio test says that if a series has nonzero terms eventually and the limit of the ratios of successive absolute terms is less than \(1\), then the series converges absolutely. Here the limit is \(0\), so the test applies for every nonzero \(c\). For \(c=0\), convergence is immediate because every term is zero.

The same conclusion can be seen directly by bounding the tail. This argument also explains why a small successive-term ratio controls all the terms that follow it.

1
Choose a point in the tail.
Because \(|c|/(n+1)^p\) tends to zero, there is an integer \(N\) such that \(|c|/(n+1)^p\le 1/2\) for every \(n\ge N\).
2
Bound each later term.
For every \(n\ge N\), the ratio formula gives \(|a_{n+1}|\le |a_n|/2\). Repeating this inequality yields \(|a_{N+k}|\le |a_N|/2^k\) for every integer \(k\ge0\).
3
Compare the tail with a geometric series.
The sum of the absolute values from \(N\) onward is at most \(|a_N|\sum_{k=0}^{\infty}(1/2)^k=2|a_N|\), which is finite. Adding the finitely many earlier absolute values still gives a finite total.

This proves absolute convergence directly. It also shows that no matter how large \(|c|\) is, the ratio eventually becomes at most \(1/2\). A larger base may postpone that point, but it does not change the limiting ratio.

Worked Examples

Worked Example: A Positive Base and a First-Power Factorial

Classify

$$ \sum_{n=1}^{\infty}\frac{5^n}{n!}. $$

Set \(a_n=5^n/n!\). Both terms in the ratio are positive, and \((n+1)!=(n+1)n!\), so

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

Taking the limit gives

$$ \lim_{n\to\infty}\frac{|a_{n+1}|}{|a_n|} = \lim_{n\to\infty}\frac{5}{n+1} =0<1. $$

The ratio test therefore proves that the series converges absolutely. As a check, the formula gives the first ratio \(|a_2|/|a_1|=5/2\): indeed, \(a_1=5\) and \(a_2=25/2\), so \((25/2)/5=5/2\), as required.

Worked Example: A Negative Base and a Fractional Factorial Power

Classify

$$ \sum_{n=1}^{\infty}\frac{(-2)^n}{(n!)^{1/2}}. $$

The terms alternate in sign, so use absolute values to test absolute convergence. With \(a_n=(-2)^n/(n!)^{1/2}\),

$$ \frac{|a_{n+1}|}{|a_n|} = \frac{2^{n+1}}{((n+1)!)^{1/2}} \cdot \frac{(n!)^{1/2}}{2^n} = \frac{2}{(n+1)^{1/2}}. $$

Since \((n+1)^{1/2}\) tends to infinity,

$$ \lim_{n\to\infty}\frac{2}{(n+1)^{1/2}}=0<1. $$

The series converges absolutely, and therefore converges. To check the calculation at \(n=1\), \(a_1=-2\) and \(a_2=4/\sqrt{2}\). Thus \(|a_2|/|a_1|=(4/\sqrt{2})/2=\sqrt{2}\), while the ratio formula gives \(2/\sqrt{2}=\sqrt{2}\), in agreement.

Worked Example: A Larger Base with a Squared Factorial

Classify

$$ \sum_{n=1}^{\infty}\frac{(-3)^n}{(n!)^2}. $$

For \(a_n=(-3)^n/(n!)^2\), the absolute ratio is

$$ \frac{|a_{n+1}|}{|a_n|} = \frac{3^{n+1}}{((n+1)!)^2} \cdot \frac{(n!)^2}{3^n} = \frac{3}{(n+1)^2}. $$

Therefore,

$$ \lim_{n\to\infty}\frac{3}{(n+1)^2}=0<1, $$

so the series converges absolutely. For a concrete tail bound, if \(n\ge2\), then \(n+1\ge3\) and \(3/(n+1)^2\le1/3\). Thus each absolute term after \(|a_2|\) is at most one third of the preceding term. Checking at \(n=2\), the ratio formula gives \(3/9=1/3\); directly, \(|a_2|=9/4\) and \(|a_3|=27/36=3/4\), so \(|a_3|/|a_2|=(3/4)/(9/4)=1/3\).

What the Sign of the Base Changes

The sign of \(c\) affects the signs of the terms, but not the absolute ratio. If \(c>0\), all terms are positive. If \(c<0\), the terms alternate in sign. In either case, the absolute terms have ratio \(|c|/(n+1)^p\), which tends to zero. Thus the conclusion is absolute convergence, not merely convergence arising from cancellation between positive and negative terms.

If \(c=0\), then \(c^n=0\) for every \(n\ge1\), so the series is the zero series. The ratio calculation is unnecessary in that case because ratios of its zero terms are undefined. Treating this case separately avoids using the ratio test where its ratio is not defined.

A Common Ratio Calculation Error

A common mistake is to forget that increasing the index changes the factorial. For example, for \(a_n=3^n/(n!)^2\), one might incorrectly claim that the ratio is \(3\), as if the denominator stayed unchanged. The incorrect calculation would suggest that the terms grow by a fixed factor of \(3\).

$$ \text{Incorrect:}\qquad \frac{a_{n+1}}{a_n}=3. $$

But \((n+1)!=(n+1)n!\), and the factorial is squared. Keeping both factors gives

$$ \frac{a_{n+1}}{a_n} = \frac{3^{n+1}}{((n+1)!)^2}\cdot\frac{(n!)^2}{3^n} = \frac{3}{(n+1)^2}. $$

The correction matters: the actual ratio tends to zero, rather than remaining at \(3\). Always expand the factorial at the new index before simplifying the ratio.

Why This Classification Is Useful

The result gives an immediate classification for an entire family, rather than requiring separate tests for different fixed bases. The base may be large, the power \(p\) may be a positive fraction, and the terms may alternate in sign; as long as \(c\) is fixed and \(p>0\), the absolute successive-term ratio tends to zero.

The conditions matter. The argument uses both that \(c\) is fixed and that \(p\) is positive. In the ratio, these assumptions appear as the fixed numerator \(|c|\) and the denominator \((n+1)^p\) tending to infinity. If those conditions change, this calculation alone does not establish the same conclusion.

Also, a ratio limit of zero describes the relative size of successive terms; it does not mean that the sum of the series is zero. The sum depends on all the terms. The ratio test classifies convergence here, while the geometric-tail argument explains why the absolute tail is finite.

Check Your Understanding

Use the successive-term ratio and state whether each series converges absolutely.

  1. For fixed \(c=7\) and \(p=1/2\), what is the ratio of successive absolute terms, and what is its limit?
  2. Why is it necessary to handle \(c=0\) separately when applying the ratio test?
  3. For \(c=-4\) and \(p=2\), does the sign alternation change the absolute-convergence classification? Explain briefly.
  4. What assumption on \(p\) ensures that \((n+1)^p\) tends to infinity?
  5. Explain how a ratio bound of \(1/2\) from some index onward gives a finite bound for the absolute tail.

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