Why Factorials Call for Successive-Term Ratios
Factorials grow rapidly: \(n!\) is the product of the positive integers from \(1\) through \(n\). Powers such as \(n^n\) also grow quickly, and comparing their growth by inspection can be difficult. In a series, however, the relationship between consecutive terms often simplifies: \((n+1)!=(n+1)n!\), while the powers can be handled by quotient rules. The ratio test turns that simplification into a convergence decision.
For an alternating series, the signs change, but absolute convergence is determined by the series of term magnitudes. Thus factorial expressions such as \(n!/n^n\) and \(n^n/n!\) are natural candidates for the ratio test. The outcome can be quite different: one can produce an absolutely convergent series, while the other can make the terms fail even to approach zero.
The Ratio Test for Magnitudes
Let \(\sum_{n=1}^{\infty}u_n\) be a series, and suppose its terms are nonzero for all sufficiently large \(n\). Apply the ratio test to the magnitudes by considering the limit
- when \(L<1\), the series \(\sum |u_n|\) converges, so \(\sum u_n\) converges absolutely;
- when \(L>1\), the terms \(|u_n|\) do not tend to zero, so \(\sum u_n\) diverges;
- when \(L=1\), the ratio test gives no conclusion.
For factorial terms, the essential cancellation is \( (n+1)!/n!=n+1\). It is important to retain the full power when simplifying terms such as \((n+1)^{n+1}\). The ratio is a quotient of consecutive magnitudes, not a comparison of the two formulas by looking only at their visible factors.
Why the Ratio Test Works
The test follows from comparison with a geometric series. If \(L<1\), choose a number \(q\) with \(L<q<1\). By the definition of the limit, there is an index \(N\) such that for every \(n\geq N\), \(|u_{n+1}|/|u_n|<q\). Repeatedly applying this inequality bounds every later magnitude by a geometric sequence.
For \(n\geq N\), the ratio condition gives \(|u_{n+1}|<q|u_n|\).
For each integer \(k\geq 0\), repeated application gives \(|u_{N+k}|\leq |u_N|q^k\).
The geometric series \(\sum_{k=0}^{\infty}|u_N|q^k\) converges because \(0<q<1\). Therefore the tail of \(\sum |u_n|\) converges, and adding its finitely many initial terms preserves convergence.
If \(L>1\), choose \(q\) with \(1<q<L\). Eventually, \(|u_{n+1}|/|u_n|>q\), so the magnitudes grow by at least a factor \(q\) at every subsequent step. In particular, they cannot tend to zero. A necessary condition for a series to converge is that its terms tend to zero, so the series diverges. If \(L=1\), neither argument applies; the ratio test alone cannot decide convergence.
Worked Example: An Alternating Series with \(n!/n^n\)
Worked Example: An Alternating Series with \(n!/n^n\)
Classify the series
Set \(a_n=n!/n^n\), which is positive for \(n\geq 1\). The series converges absolutely exactly when \(\sum a_n\) converges. Compute the ratio of consecutive magnitudes:
Using the standard limit \(\lim_{n\to\infty}(1+1/n)^n=e\), the ratio limit is
The ratio test says \(\sum a_n\) converges. Therefore the original alternating series converges absolutely. As a check, for \(n=1\), \(a_1=1!/1^1=1\), and for \(n=2\), \(a_2=2!/2^2=1/2\); their ratio is \(a_2/a_1=1/2\), agreeing with the general ratio formula at \(n=1\).
Worked Example: An Alternating Series with \(n^n/n!\)
Worked Example: An Alternating Series with \(n^n/n!\)
Consider
Write \(b_n=n^n/n!\). Its successive-term ratio is
Consequently,
In fact, the ratio is eventually greater than a fixed number larger than \(1\), so the magnitudes grow rather than approach zero. The terms of the signed series also fail to tend to zero, since their absolute values are \(b_n\). The necessary term condition for convergence is violated, and the alternating series diverges. Alternating signs do not repair terms whose magnitudes fail to approach zero.
As a numerical check on the ratio formula, \(b_1=1^1/1!=1\), \(b_2=2^2/2!=2\), and \(b_2/b_1=2\). The formula gives \((1+1/1)^1=2\), as required.
Worked Example: Adding an Exponential Factor
Worked Example: Adding an Exponential Factor
Classify
Let \(c_n=n!/(2^n n^n)\). Then
The limit is \(L=1/(2e)<1\). The ratio test therefore proves that \(\sum c_n\) converges, and the alternating series converges absolutely. For a direct check at the first step, \(c_1=1/2\) and \(c_2=2!/(2^2\cdot 2^2)=1/8\), so \(c_2/c_1=1/4\). Substituting \(n=1\) into the ratio formula gives \(\frac12(1/2)^1=1/4\), in agreement.
What the Classification Says—and Does Not Say
The ratio test classifies the series of magnitudes when its limit is less than \(1\). That conclusion is stronger than merely knowing that the signs alternate: it establishes absolute convergence. If the ratio limit is greater than \(1\), the failure of the terms to approach zero is enough to establish divergence, regardless of the sign pattern.
A limit of \(1\) is different. It does not imply convergence, and it does not imply divergence. For example, the positive series \(\sum 1/n\) diverges, while \(\sum 1/n^2\) converges; in both cases the ratio of consecutive terms tends to \(1\). In an alternating problem, other results may be relevant when the ratio test is inconclusive, but the ratio limit by itself settles neither absolute nor conditional convergence.
A Common Simplification Error
When calculating the ratio for \(a_n=n!/n^n\), a tempting but incorrect simplification is to treat the power \((n+1)^{n+1}\) as though it were \((n+1)n^n\). This produces the wrong calculation
and might lead to the false claim that the ratio test is inconclusive. The exponent rule does not permit that replacement: \((n+1)^{n+1}=(n+1)(n+1)^n\), not \((n+1)n^n\). Keeping the full power gives the correct simplification
The practical safeguard is to write out the factorial cancellation and the power quotient separately before taking a limit.
Check Your Understanding
For each question, show the successive-term ratio before stating what the test establishes.
- For \(\sum_{n=1}^{\infty}(-1)^{n-1}n!/(3^n n^n)\), find the ratio limit and classify the series.
- For \(\sum_{n=1}^{\infty}(-1)^{n-1}n^n/(2^n n!)\), determine whether the terms approach zero and classify the series.
- State what the ratio test concludes if the ratio limit for a magnitude series is \(1\).
- Explain why absolute convergence of an alternating series is decided by its magnitudes rather than its signs.
- In simplifying a factorial ratio, what identity allows \((n+1)!\) to cancel against \(n!\)?
