Why Shifted Factorials Need Careful Simplification
Factorials can make a series term look more complicated than it is. When the factorials have indices that differ by a fixed number, most of their factors are shared. Cancelling those shared factors reveals a finite product, and that product behaves like a power of \(n\) for large \(n\).
That power is often enough to determine whether a series converges. The important steps are to simplify the factorial quotient exactly, identify the resulting power, and compare the terms with a familiar power series. Taking a limit before cancelling can hide this structure.
Cancel the Shared Factors
For a nonnegative integer \(k\), \(k!\) denotes the product of the positive integers from \(1\) through \(k\), with \(0!=1\). Suppose \(r\) and \(s\) are fixed integers, and \(n\) is large enough that both \(n+r\) and \(n+s\) are nonnegative.
If \(r>s\), expanding the larger factorial shows that every factor in \((n+s)!\) cancels, leaving the factors above \(n+s\). If \(r<s\), the same cancellation leaves a product in the denominator.
The number of factors left is the absolute difference between the shifts. These are exact identities, not approximations. For example, if \(r<s\), there are \(s-r\) factors in the denominator. Each factor has the form \(n+c\), where \(c\) is fixed.
Worked Example: Cancel a Quotient with More Factors Below
Simplify \((n+2)!/(n+5)!\). Expanding the denominator gives
Cancel the common factor \((n+2)!\):
To check the cancellation, multiply the simplified expression by \((n+5)!\):
The result agrees with the numerator, so the simplification is correct.
The Power Hidden in a Shifted Ratio
The exact product also gives a precise large-\(n\) description. Factor \(n\) out of every remaining term. Each factor \(n+c\) becomes \(n(1+c/n)\). The fixed constants \(c\) do not affect the limiting power, because \(1+c/n\) tends to \(1\).
Here “behaves like” means that the quotient of the two expressions tends to \(1\). A proof follows directly from the finite product left after cancellation.
Put \(m=s-r\). Cancellation gives a denominator with \(m\) factors: $$ \frac{(n+r)!}{(n+s)!} = \frac{1}{(n+r+1)\cdots(n+s)}. $$
There are \(m\) factors, so $$ n^m\frac{(n+r)!}{(n+s)!} = \frac{1}{(1+(r+1)/n)\cdots(1+s/n)}. $$
Every factor in the denominator tends to \(1\), and there are only finitely many factors. The displayed expression therefore tends to \(1\). Since \(m=s-r\), this is the claimed limit.
If \(r=s\), the factorial quotient is exactly \(1\). If \(r>s\), cancellation leaves \(r-s\) numerator factors. Factoring \(n\) out of each shows that division by \(n^{r-s}\) leaves a finite product whose factors tend to \(1\). This gives the same stated limit.
This argument explains why only the shift difference matters for the leading power. The individual shifts affect lower-order factors, but not the power that governs the eventual size of the quotient.
Using the Power to Classify a Series
For a series with positive terms, the limit comparison test can turn this power estimate into a convergence conclusion. In the form used here, if \(a_n>0\), \(b_n>0\), and \(a_n/b_n\) tends to a finite positive number, then \(\sum a_n\) and \(\sum b_n\) either both converge or both diverge. The \(p\)-series \(\sum 1/n^p\) converges when \(p>1\) and diverges when \(p\leq 1\).
Consider terms that include both a shifted factorial ratio and a power \(n^{-q}\), where \(q\) is fixed. The theorem shows that their size is governed by \(n^{r-s-q}\), or equivalently by \(1/n^{q+s-r}\). Thus the decisive exponent is \(q+s-r\).
This classification follows because the ratio of the general term to \(1/n^{q+s-r}\) tends to \(1\). The limit comparison test therefore reduces the decision to the \(p\)-series criterion.
Worked Example: A Convergent Quotient with Three Denominator Factors
Classify the series
First cancel the common factorial:
Compare with \(b_n=1/n^3\). The ratio is
As \(n\) tends to infinity, each factor \(1+c/n\) tends to \(1\), so \(a_n/b_n\) tends to \(1\). The comparison series \(\sum 1/n^3\) converges because \(3>1\). The limit comparison test therefore gives convergence of the factorial series.
As a check on the cancellation, multiplying \(a_n\) by \((n+5)!\) gives $$ \frac{(n+5)!}{(n+3)(n+4)(n+5)} = (n+2)!, $$ as required.
Worked Example: The Boundary Case with Two Denominator Factors
Classify
Cancel the common factorial factors:
Compare with \(1/n^2\). The ratio of the terms is
Since \(\sum 1/n^2\) converges, the given series converges. The two leftover denominator factors produce decay like \(1/n^2\), which is beyond the convergence boundary \(1/n\).
For a direct check of the algebra, multiplying the simplified term by \((n+3)!\) gives $$ \frac{(n+3)!}{(n+2)(n+3)} = \frac{(n+3)(n+2)(n+1)!}{(n+2)(n+3)} = (n+1)!. $$
Worked Example: One Denominator Factor Gives Divergence
Consider
Cancellation leaves one factor in the denominator:
To compare with \(1/n\), compute
The harmonic series \(\sum 1/n\) diverges, so limit comparison shows that this series diverges as well. The shift by \(2\) changes only finitely many initial terms and the location of the terms, not the harmonic rate of decay.
Check the cancellation by multiplying \(1/(n+2)\) by \((n+2)!\): $$ \frac{(n+2)!}{n+2}=(n+1)!, $$ which recovers the original numerator.
Worked Example: A Polynomial Factor Changes the Boundary
Classify the series
After cancelling, three factors remain in the denominator:
Compare this term with \(1/n\). The ratio is
The harmonic series diverges, so the given series diverges by limit comparison. The factor \(n^2\) uses up two of the three powers supplied by the denominator; the remaining decay is only like \(1/n\).
As an algebra check, multiplying the simplified term by \((n+4)!\) gives $$ \frac{n^2(n+4)!}{(n+2)(n+3)(n+4)} = n^2(n+1)!, $$ which is exactly the numerator in the original term.
Common Pitfall: Treating Factorials as Whole Symbols
A factorial is a product, not an indivisible symbol. Cancelling the factorials does not mean cancelling their indices, nor does it mean that the quotient is a constant. For instance, the following calculation is wrong:
The shift difference is \(3\), but the three factors left after cancellation depend on \(n\). The correct calculation is
There is another common trap: applying the Ratio Test to these terms and stopping when the limit is \(1\). For \(a_n=1/(n+2)\), for example,
A limit of \(1\) makes the Ratio Test inconclusive; it does not establish convergence. In this case, cancellation and comparison reveal the harmonic rate, so the series diverges. Simplifying first makes the relevant power visible.
Summary: Read the Shift Difference
For fixed shifts, the difference \(s-r\) gives the number of factors left in the denominator when \(r<s\). With no additional power of \(n\), a denominator gap greater than \(1\) gives convergence by comparison with \(1/n^p\); a gap of \(1\) gives harmonic behavior and divergence. A quotient with \(r\ge s\) does not even tend to zero, so its series diverges by the term test.
When an extra factor \(n^{-q}\) is present, combine its power with the shift difference. The resulting \(p\)-series exponent is \(q+s-r\). Convergence occurs when that exponent is greater than \(1\); at \(1\), the harmonic boundary still diverges. In every case, the reliable procedure is the same: cancel exactly, identify the power, and then use a comparison whose conditions are satisfied.
Check Your Understanding
For each question, simplify the factorial quotient before deciding what the remaining factors imply.
- Simplify \((n+3)!/(n+6)!\) as a product.
- What power of \(n\) describes \((n+1)!/(n+5)!\) for large \(n\)?
- Does \(\sum_{n=1}^{\infty}(n+2)!/(n+4)!\) converge or diverge? Identify the comparison series.
- Classify \(\sum_{n=1}^{\infty} n(n+1)!/(n+4)!\), and state the limiting comparison ratio.
- Why does a successive-term ratio tending to \(1\) fail to decide convergence?
