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Ratio, Root, and Test Selection · Tutorial 680 of 998

Products and Quotients of Factorials

Use exact factorial cancellation to evaluate ratio-test limits and determine when series built from factorial products converge.

Advanced 9 min read

What You'll Learn

  • Form the ratio of consecutive terms containing several factorials.
  • Simplify factorial quotients by expanding only the factors that change.
  • Evaluate ratios involving factorials such as (an)! and (bn)!.
  • Predict ratio-test limits from the slopes of factorial arguments.
  • Apply the ratio test and recognize when its limit is inconclusive.
  • Avoid cancellation errors by tracking every factor.

Why Factorial Ratios Need Care

When a series term contains several factorials, writing the next term and dividing can produce a long expression. The key is to use the definition of factorial to identify exactly which factors change from one index to the next. Careful cancellation then often turns a complicated-looking quotient into a limit that is straightforward to evaluate.

This method is especially useful for factorials such as \((an)!\), where \(a\) is a positive integer. Advancing the index from \(n\) to \(n+1\) adds \(a\) factors—not just one. Keeping track of all of them is essential to getting the correct ratio-test limit.

Definition: For a positive sequence \((u_n)\), the ratio of successive terms is \(\frac{u_{n+1}}{u_n}\). If its limit is \(L<1\), the series \(\sum u_n\) converges. If \(L>1\), or if the ratio tends to infinity, the series diverges. If \(L=1\), the ratio test gives no conclusion.

The ratio test also applies to a series whose terms may have signs by using \(\left|\frac{u_{n+1}}{u_n}\right|\). The examples here have positive terms, so the absolute-value bars are unnecessary. The test requires the successive-term ratio to be defined for all sufficiently large \(n\), as it is for the positive factorial terms considered below.

The Exact Cancellation Rule

The factorial identity \(m!=1\cdot2\cdots m\) gives a useful rule. If \(r\) is a positive integer, then

$$ \frac{(m+r)!}{m!}=(m+1)(m+2)\cdots(m+r). $$

Only the factors added between \(m!\) and \((m+r)!\) remain. For a factorial whose argument is \(an+c\), advancing \(n\) by one increases the argument by \(a\). Thus the quotient involves exactly \(a\) factors.

Rule: For positive integers \(a\) and nonnegative integers \(c\), $$ \frac{(a(n+1)+c)!}{(an+c)!} =(an+c+1)(an+c+2)\cdots(an+c+a). $$ Expand each changing factorial by this rule, then cancel common factors before taking a limit.

This is exact cancellation, not an approximation. In particular, a factorial with an argument that increases by \(a\) contributes \(a\) consecutive factors to the ratio. Omitting even one of them can change the limit and the conclusion about the series.

Worked Example: A Squared Factorial over a Doubled Index

Worked Example: A Squared Factorial over a Doubled Index

Classify the series with terms \(u_n=\frac{(n!)^2}{(2n)!}\), for \(n\geq1\).

Write the next term and divide:

$$ \frac{u_{n+1}}{u_n} = \frac{((n+1)!)^2}{(2n+2)!} \frac{(2n)!}{(n!)^2}. $$

Use \((n+1)!/n!=n+1\) and \((2n+2)!/(2n)!=(2n+1)(2n+2)\):

$$ \frac{u_{n+1}}{u_n} = \frac{(n+1)^2}{(2n+1)(2n+2)} = \frac{(n+1)^2}{(2n+1)\,2(n+1)} = \frac{n+1}{2(2n+1)}. $$

Therefore,

$$ L=\lim_{n\to\infty}\frac{n+1}{2(2n+1)} =\frac{1}{4}. $$

Since \(1/4<1\), the ratio test shows that \(\sum_{n=1}^{\infty}\frac{(n!)^2}{(2n)!}\) converges. As a check on the cancellation, \(u_1=1/2\) and \(u_2=4/24=1/6\), so \(u_2/u_1=1/3\). The ratio formula at \(n=1\) also gives \(2/[2(3)]=1/3\).

Factorials with Arguments \(an\) and \(bn\)

Consider the single quotient \(u_n=\frac{(an)!}{(bn)!}\), where \(a\) and \(b\) are fixed positive integers. The successive-term ratio is

$$ \frac{u_{n+1}}{u_n} = \frac{(an+a)!}{(an)!} \frac{(bn)!}{(bn+b)!} = \frac{(an+1)\cdots(an+a)} {(bn+1)\cdots(bn+b)}. $$

The numerator has \(a\) factors and the denominator has \(b\). When the ratio is divided through by the appropriate power of \(n\), each factor in the numerator contributes a leading coefficient \(a\), and each denominator factor contributes \(b\). This gives the following limit behavior:

$$ \frac{u_{n+1}}{u_n} \sim \frac{a^a}{b^b}n^{a-b}. $$

Here, \(\sim\) means that the quotient of the two sides tends to \(1\). If \(a<b\), the ratio tends to \(0\), so the series of these positive terms converges by the ratio test. If \(a>b\), the ratio tends to infinity, so the series diverges. If \(a=b\), then \(u_n=1\) for every \(n\), and the series diverges. The factorial quotient alone therefore gives different behavior depending on the two index multipliers.

Worked Example: A Factorial Quotient with Different Multipliers

Classify the series with terms \(u_n=\frac{(2n)!}{(3n)!}\), for \(n\geq1\).

Advancing \(n\) by one adds two factors to the numerator factorial and three to the denominator factorial:

$$ \frac{u_{n+1}}{u_n} = \frac{(2n+1)(2n+2)} {(3n+1)(3n+2)(3n+3)}. $$

The numerator has degree two in \(n\), and the denominator has degree three. More explicitly, dividing the numerator and denominator by \(n^3\) shows that the ratio tends to \(0\). Thus the ratio test gives convergence of \(\sum_{n=1}^{\infty}\frac{(2n)!}{(3n)!}\).

Check the exact expression at \(n=1\): it gives \((3)(4)/[(4)(5)(6)]=1/10\). Directly, \(u_1=2/6=1/3\) and \(u_2=24/720=1/30\), so \(u_2/u_1=(1/30)/(1/3)=1/10\), as required.

A General Limit for Products of Factorials

The same reasoning handles any fixed finite product of factorials whose arguments are integer multiples of \(n\), with fixed shifts. Let

$$ T_n= \frac{\prod_{i=1}^{p}(a_i n+c_i)!} {\prod_{j=1}^{q}(b_j n+d_j)!}, $$

where all \(a_i\) and \(b_j\) are positive integers and all \(c_i\) and \(d_j\) are fixed nonnegative integers. A constant positive multiplier on \(T_n\) would cancel from its successive-term ratio. For the factorials shown, applying the exact cancellation rule gives

$$ \frac{T_{n+1}}{T_n} = \frac{ \prod_{i=1}^{p}\prod_{r=1}^{a_i}(a_i n+c_i+r) }{ \prod_{j=1}^{q}\prod_{s=1}^{b_j}(b_j n+d_j+s) }. $$

Each factor in the numerator is linear in \(n\), and there are \(\sum_{i=1}^{p}a_i\) such factors. The denominator has \(\sum_{j=1}^{q}b_j\) linear factors. Their leading coefficients give the limit behavior below.

Theorem: Define $$ \Delta=\sum_{i=1}^{p}a_i-\sum_{j=1}^{q}b_j, \qquad K=\frac{\prod_{i=1}^{p}a_i^{a_i}}{\prod_{j=1}^{q}b_j^{b_j}}. $$ Then $$ \frac{T_{n+1}}{T_n}\sim K n^{\Delta}. $$ If \(\Delta<0\), the ratio tends to \(0\); if \(\Delta>0\), it tends to infinity. If \(\Delta=0\), the ratio tends to \(K\). Consequently, when \(\Delta=0\), the ratio test proves convergence if \(K<1\), and divergence if \(K>1\). If \(K=1\), it is inconclusive.

Why the General Limit Holds

For each numerator factor \(a_i n+c_i+r\), divide by \(n\). Its quotient tends to \(a_i\), since \(a_i+(c_i+r)/n\) tends to \(a_i\). There are exactly \(a_i\) factors for that index \(i\), so their combined leading contribution is \(a_i^{a_i}n^{a_i}\). Multiplying across the numerator gives the leading contribution \(\left(\prod_i a_i^{a_i}\right)n^{\sum_i a_i}\). Applying the same reasoning to the denominator gives \(\left(\prod_j b_j^{b_j}\right)n^{\sum_j b_j}\). Dividing these expressions yields \(Kn^\Delta\), and the quotient of the exact ratio by \(Kn^\Delta\) tends to \(1\). This proves the stated asymptotic formula.

The formula is useful as a prediction, but the exact ratio should still be formed first. Exact cancellation can make the limit simpler and helps prevent mistakes with shifts or repeated factorials.

Worked Example: A Product with Three Factorials

Classify the series with terms \(u_n=\frac{((2n)!)^2}{n!(3n)!}\), for \(n\geq1\).

Write the successive-term ratio, expanding each factorial by the number of factors it gains:

$$ \frac{u_{n+1}}{u_n} = \frac{[(2n+1)(2n+2)]^2} {(n+1)(3n+1)(3n+2)(3n+3)}. $$

Use \(2n+2=2(n+1)\) and \(3n+3=3(n+1)\):

$$ \frac{u_{n+1}}{u_n} = \frac{4(n+1)^2(2n+1)^2} {3(n+1)^2(3n+1)(3n+2)} = \frac{4(2n+1)^2}{3(3n+1)(3n+2)}. $$

Thus,

$$ L=\lim_{n\to\infty} \frac{4(2n+1)^2}{3(3n+1)(3n+2)} = \frac{16}{27}. $$

Since \(16/27<1\), the series converges. The general formula agrees: the numerator slopes are \(2\) and \(2\), and the denominator slopes are \(1\) and \(3\), so \(\Delta=0\) and \(K=(2^2\cdot2^2)/(1^1\cdot3^3)=16/27\). As a check at \(n=1\), the simplified ratio is \(4(3)^2/[3(4)(5)]=3/5\). Directly, \(u_1=4/6=2/3\), while \(u_2=24^2/(2\cdot720)=2/5\), and \(u_2/u_1=3/5\).

When the Ratio Limit Exceeds One

A limit greater than one means that successive terms eventually grow by a factor bounded away from one. In particular, the terms cannot tend to zero, which is necessary for a series to converge. This is why a ratio limit greater than one proves divergence rather than convergence.

Worked Example: Reversing the Factorial Quotient

Classify the series with terms \(v_n=\frac{(2n)!}{(n!)^2}\), for \(n\geq1\).

Its successive-term ratio is

$$ \frac{v_{n+1}}{v_n} = \frac{(2n+1)(2n+2)}{(n+1)^2} = \frac{(2n+1)\,2(n+1)}{(n+1)^2} = \frac{2(2n+1)}{n+1}. $$

Therefore \(L=4\). Since \(4>1\), the ratio test shows that \(\sum_{n=1}^{\infty}\frac{(2n)!}{(n!)^2}\) diverges. Indeed, \(v_1=2\), \(v_2=6\), and \(v_2/v_1=3\), which agrees with the ratio formula at \(n=1\): \(2(3)/2=3\).

A Common Cancellation Mistake

A frequent error is to treat \((2n+2)!/(2n)!\) as though it contributed only the factor \(2n+2\). It contributes two factors, because the argument has increased by two:

$$ \frac{(2n+2)!}{(2n)!}=(2n+1)(2n+2). $$

For the first worked example, omitting \(2n+1\) would incorrectly give

$$ \frac{(n+1)^2}{2n+2}=\frac{n+1}{2}, $$

which does not even have the correct limit. The correct ratio is \(\frac{(n+1)^2}{(2n+1)(2n+2)}\), with limit \(1/4\). At \(n=1\), the incorrect expression gives \(1\), while the actual ratio \(u_2/u_1\) is \(1/3\). Expanding every changed factorial before canceling prevents this error.

Choosing the Right Conclusion

After calculating the limit, apply the ratio test precisely. A limit below one proves convergence; a limit above one proves divergence. A limit equal to one does not settle the question, even if the exact ratio has been simplified correctly. In that boundary case, another argument is needed. The general slope-and-leading-coefficient formula is a way to organize the ratio calculation, not a replacement for the ratio test's conditions.

The method matters because factorial expressions can hide a simple balance: the number of factors contributed by each argument controls the power of \(n\) in the successive-term ratio, while the multipliers control its leading constant. Keeping those two pieces separate makes it easier to see whether the ratio tends to zero, to a value below or above one, or to the inconclusive boundary value one.

Check Your Understanding

Use exact cancellation to form each ratio before deciding what the ratio test says.

  1. For \(u_n=\frac{(n!)^3}{(3n)!}\), find \(\lim_{n\to\infty}u_{n+1}/u_n\) and classify the series.
  2. For \(v_n=\frac{(3n)!}{(2n)!}\), find the successive-term ratio and determine whether the ratio test proves convergence or divergence.
  3. For \(w_n=\frac{((2n)!)^2}{(n!)^2(4n)!}\), use the general slope formula to find \(\Delta\) and \(K\), then state the ratio-test conclusion.
  4. Why does \((3n+3)!/(3n)!\) contain three factors rather than one?
  5. What conclusion can the ratio test give when the successive-term ratio tends to \(1\)?

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