Why Successive-Term Ratios Help
Factorials and finite products can make a series term look complicated. But comparing consecutive terms often removes most of that complexity: the ratio between \(a_{n+1}\) and \(a_n\) retains only the factors that changed. This is the central pattern behind applying the Ratio Test to these series.
The important work is not merely forming a quotient. It is simplifying it correctly. Factorials such as \((2n)!\) add more than one factor when \(n\) increases by one, while a product whose upper limit changes adds exactly one new factor. Once the ratio is simplified, its limit gives a convergence conclusion only under the Ratio Test’s conditions.
The Ratio Test and Its Conditions
Suppose \(a_n\) is nonzero for all sufficiently large \(n\), and the limit below exists:
The absolute value matters for series whose terms may be negative: the test classifies absolute convergence when \(L<1\). For positive-term series, the absolute value does not change the ratio. In either case, the quotient must be defined eventually; a finite number of initial terms do not affect convergence.
Here is why the test works. If \(L<1\), choose a number \(q\) with \(L<q<1\). By the definition of the limit, for all sufficiently large \(n\), \(\left|\frac{a_{n+1}}{a_n}\right|<q\). Repeatedly applying this inequality bounds the later terms by a constant times the terms of a convergent geometric series. If \(L>1\), choose \(q\) with \(1<q<L\). Eventually \(|a_{n+1}|>q|a_n|\), so the terms do not approach zero. The series therefore diverges. When \(L=1\), neither comparison follows: the ratio may be one for a divergent series or may approach one for a convergent series.
Canceling Factorials Carefully
For a positive integer \(m\), \(m!=m(m-1)\cdots 2\cdot1\). When \(n\) is replaced by \(n+1\), expand just enough of each factorial to reveal the factors that differ. For example, \((n+1)!=(n+1)n!\), while \((2n+2)!=(2n+2)(2n+1)(2n)!\). The second identity has two new factors because its argument increases by two.
In a quotient of factorials, write the new factors in the numerator and denominator separately, then cancel the unchanged factorial blocks. Avoid canceling pieces that are not common factors of the entire numerator and denominator.
Worked Example: A Quotient of Two Factorials
Determine whether the series with terms \(a_n=\frac{(n!)^2}{(2n)!}\), for \(n\ge1\), converges.
First write the next term and form the ratio:
Use \((n+1)!=(n+1)n!\) and \((2n+2)!=(2n+2)(2n+1)(2n)!\):
Since \(2n+2=2(n+1)\), cancel one factor of \(n+1\):
Because \(L=\frac14<1\), the Ratio Test says \(\sum_{n=1}^{\infty}\frac{(n!)^2}{(2n)!}\) converges absolutely. The terms are positive, so this is convergence of the series.
Check: \(a_1=\frac12\) and \(a_2=\frac16\), so \(\frac{a_2}{a_1}=\frac13\). The derived ratio at \(n=1\) is \(\frac{2}{2(3)}=\frac13\), as required.
Factorials Combined with Exponentials
A fixed exponential factor is often simple to handle: if a term contains \(c^n\), replacing \(n\) by \(n+1\) contributes the factor \(c\). The factorial quotient contributes its own changing factors. Multiply those pieces together before taking the limit.
Worked Example: An Exponential and a Factorial Quotient
Test the series with terms \(a_n=\frac{3^n n!}{(2n)!}\), for \(n\ge1\).
Form the next-term ratio, using \(3^{n+1}/3^n=3\), \((n+1)!/n!=n+1\), and the two new factors in \((2n+2)!\):
Now use \(2n+2=2(n+1)\) and cancel:
Since \(0<1\), the Ratio Test proves that the series converges absolutely.
Check: \(a_1=\frac32\) and \(a_2=\frac34\), giving \(\frac{a_2}{a_1}=\frac12\). Substitution of \(n=1\) into the simplified ratio gives \(\frac{3}{2(3)}=\frac12\).
Finite Products: Keep the Newly Added Factor
Let \(P_n=\prod_{k=1}^{n}u_k\), where the factors are defined and nonzero. Then \(P_{n+1}\) consists of the factors of \(P_n\), together with one additional factor \(u_{n+1}\). Consequently, \(P_{n+1}/P_n=u_{n+1}\). This cancellation is exact; it does not require expanding the entire product.
If a series term also includes another factor, such as \(r^n\), include its ratio as well. The product’s last factor and the exponential’s ratio together give the successive-term ratio.
Worked Example: A Product with a Geometric Factor
For \(n\ge1\), define \(a_n=\left(\frac12\right)^n\prod_{k=1}^{n}\frac{k}{k+3}\). Determine whether \(\sum a_n\) converges.
When \(n\) increases to \(n+1\), the exponential contributes \(\frac12\), and the product gains the factor \(\frac{n+1}{n+4}\). Therefore,
Taking the limit gives
Since \(L=\frac12<1\), the series converges by the Ratio Test.
The product can also be checked by writing out its factors:
Thus \(a_1=\frac18\), and \(a_2=\frac1{40}\), so \(\frac{a_2}{a_1}=\frac15\). The ratio formula at \(n=1\) gives \(\frac12\cdot\frac25=\frac15\), confirming the cancellation and the result.
When the Ratio Limit Is One
A common mistake is to treat \(L=1\) as evidence for divergence or convergence. It is neither. The Ratio Test simply cannot decide in that case. A product may collapse to a familiar term while its successive-term ratio approaches one, so the ratio itself does not reveal enough about the series.
Worked Example: A Product Ratio That Does Not Decide
Consider \(b_n=\prod_{k=1}^{n}\frac{k}{k+1}\), for \(n\ge1\). Consecutive factors cancel:
From the newly added factor,
The Ratio Test is inconclusive. In this example the series is \(\sum_{n=1}^{\infty}\frac1{n+1}\), which diverges; that fact must be established by another method, not inferred from \(L=1\).
Check: \(b_1=\frac12\) and \(b_2=\frac13\), so \(\frac{b_2}{b_1}=\frac23\). The ratio formula at \(n=1\) also gives \(\frac23\).
A Cancellation Pitfall
Here is a concrete error to avoid. In the first factorial example, one might write \((2n+2)!/(2n)!=(2n+2)\), forgetting the factor \((2n+1)\). That would produce an incorrect ratio:
This expression grows without bound, which would incorrectly suggest divergence. The correct expansion is \((2n+2)!=(2n+2)(2n+1)(2n)!\), so the correct ratio is \(\frac{(n+1)^2}{(2n+2)(2n+1)}\), with limit \(\frac14\). Always count how many integers are added to a factorial’s argument: increasing \(2n\) to \(2n+2\) adds two factors, not one.
A Reliable Ratio-Calculation Routine
Use the following sequence when factorials or finite products appear in a series term. Keeping the algebra and the test conclusion separate makes it easier to spot both cancellation errors and unjustified claims.
Replace every \(n\) by \(n+1\), including upper limits in products.
Calculate \(\left|\frac{a_{n+1}}{a_n}\right|\), including any exponential contribution.
For factorials, show the new factors explicitly; for a finite product, keep its newly added factor.
Cancel common factors, simplify, and find \(L\).
Use the Ratio Test if \(L<1\) or \(L>1\); if \(L=1\), say that the test is inconclusive.
Check Your Understanding
For each problem, form and simplify the successive-term ratio before deciding what the Ratio Test can establish.
- For \(a_n=\frac{n!}{4^n}\), find \(\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|\). What does the Ratio Test conclude?
- For \(a_n=\frac{(n!)^2}{(3n)!}\), which new factors appear in the denominator when \(n\) is replaced by \(n+1\)?
- If \(P_n=\prod_{k=1}^{n}\frac{2k}{2k+1}\), what is \(P_{n+1}/P_n\)?
- What does the Ratio Test conclude when the ratio limit is exactly one?
- Explain why \((2n+2)!/(2n)!\) contains two factors rather than one.
