Why factorial series can simplify
Factorials grow quickly, so a term involving factorials may initially look complicated. But factorials also have a useful feature: consecutive factorials differ by one simple factor, since \((n+1)!=(n+1)n!\). That relationship can turn a factorial ratio into a difference of simpler terms. When those differences are added, many intermediate terms cancel.
The cancellation is only part of the analysis. A series is defined through the limit of its partial sums. After simplifying a finite partial sum, we must check whether the terms left at its ends have limits that make the partial sums converge to a finite number. A cancellation pattern alone does not guarantee convergence.
Factorial ratios and useful identities
For a positive integer \(n\), the factorial is \(n!=1\cdot2\cdot3\cdots n\), with \(0!=1\). If \(n\) is a nonnegative integer and \(k\) is a positive integer, then \[ (n+k)!=(n+k)(n+k-1)\cdots(n+1)n!. \] Consequently, whenever the denominator is nonzero, \[ \frac{n!}{(n+k)!}=\frac{1}{(n+1)(n+2)\cdots(n+k)}. \]
The consecutive-factorial identity also gives differences. For example, \[ \frac{1}{n!}-\frac{1}{(n+1)!} =\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!} =\frac{n}{(n+1)!}. \] Thus the term \(n/(n+1)!\) can be summed by cancellation. Another useful identity is \[ \frac{1}{(n+1)!}-\frac{1}{(n+2)!} =\frac{n+2}{(n+2)!}-\frac{1}{(n+2)!} =\frac{n+1}{(n+2)!}. \] In each identity, putting the two fractions over a common denominator verifies the numerator. This is a good habit: factorial expressions can be easy to misread, so check the proposed difference algebraically before using it.
For a difference series, write out the finite sum through an arbitrary index \(N\). That reveals which terms cancel and which remain. Only after obtaining the exact formula for \(S_N\) should we take the limit as \(N\) tends to infinity.
Worked Example: A ratio of factorials
Worked Example: A ratio of factorials
Determine whether \(\displaystyle\sum_{n=1}^{\infty}\frac{n!}{(n+2)!}\) converges, and find its sum if it does.
First expand the denominator: \[ (n+2)!=(n+2)(n+1)n!. \] Cancel the common factor \(n!\): \[ \frac{n!}{(n+2)!} =\frac{n!}{(n+2)(n+1)n!} =\frac{1}{(n+1)(n+2)}. \]
Now express this as a difference. A common denominator gives \[ \frac{1}{n+1}-\frac{1}{n+2} =\frac{(n+2)-(n+1)}{(n+1)(n+2)} =\frac{1}{(n+1)(n+2)}. \] Therefore \[ \frac{n!}{(n+2)!}=\frac{1}{n+1}-\frac{1}{n+2}. \]
For a finite upper index \(N\), the partial sum is \[ S_N=\sum_{n=1}^{N}\left(\frac{1}{n+1}-\frac{1}{n+2}\right). \] Writing out its terms shows the cancellation: \[ S_N=\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\cdots+ \left(\frac{1}{N+1}-\frac{1}{N+2}\right) =\frac12-\frac{1}{N+2}. \] Since \(1/(N+2)\) tends to zero, \(S_N\) tends to \(1/2\). The series converges to \(1/2\).
As a check, the summand at \(n=1\) is \(1!/3!=1/6\), and the difference formula gives \(1/2-1/3=1/6\). Also, the finite-sum formula at \(N=1\) gives \(1/2-1/3=1/6\), as required.
Why the cancellation works
The mechanism does not depend on factorials alone. If a term has the form \(b_n-b_{n+1}\), then adding the first \(N\) terms gives \[ (b_1-b_2)+(b_2-b_3)+\cdots+(b_N-b_{N+1}). \] Each term \(b_j\) for \(2\le j\le N\) occurs once with a positive sign and once with a negative sign. Those occurrences sum to zero, leaving only \(b_1-b_{N+1}\). This proves the finite cancellation formula, rather than relying on a pattern inferred from a few initial terms.
It follows that if \(b_N\) approaches a finite limit \(L\), then the partial sums approach \(b_1-L\). The factorial examples below use this finite identity and then check the remaining terms by taking a limit. The limit step matters: if the remaining terms do not approach values that give a finite partial-sum limit, the series does not converge.
Worked Example: Reciprocal factorial differences
Worked Example: Reciprocal factorial differences
Find the sum of \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{(n+1)!}\).
Use the factorial identity \((n+1)!=(n+1)n!\) to check the decomposition: \[ \frac{1}{n!}-\frac{1}{(n+1)!} =\frac{n+1}{(n+1)!}-\frac{1}{(n+1)!} =\frac{n}{(n+1)!}. \] Hence the \(n\)th term is a difference of reciprocal factorials.
The \(N\)th partial sum is \[ S_N=\sum_{n=1}^{N}\left(\frac{1}{n!}-\frac{1}{(n+1)!}\right) =\left(1-\frac{1}{2!}\right)+\left(\frac{1}{2!}-\frac{1}{3!}\right)+\cdots+ \left(\frac{1}{N!}-\frac{1}{(N+1)!}\right) =1-\frac{1}{(N+1)!}. \] As \(N\) tends to infinity, \((N+1)!\) grows without bound, so \(1/(N+1)!\) tends to zero. Therefore \(S_N\) tends to \(1\), and the series converges to \(1\).
Check the first term: \(1/(1+1)!=1/2\), and \(1/1!-1/2!=1-1/2=1/2\). The partial-sum formula at \(N=1\) also gives \(1-1/2!=1/2\). Both checks agree with the original series.
Worked Example: Cancellation that does not give convergence
Worked Example: Cancellation that does not give convergence
Determine whether \(\displaystyle\sum_{n=1}^{\infty}\left(\frac{(n+1)!}{n!}-\frac{n!}{(n-1)!}\right)\) converges.
Simplify the factorial ratios separately: \[ \frac{(n+1)!}{n!}=n+1 \qquad\text{and}\qquad \frac{n!}{(n-1)!}=n. \] Thus each term of the series equals \((n+1)-n=1\). Its \(N\)th partial sum is \[ S_N=\underbrace{1+1+\cdots+1}_{N\text{ terms}}=N. \] As \(N\) tends to infinity, \(S_N\) grows without bound, so the series diverges.
This calculation can also be written as a finite cancellation. Set \(b_n=n!/(n-1)!=n\). Then the \(n\)th term is \(b_{n+1}-b_n\), and \[ S_N=(b_2-b_1)+(b_3-b_2)+\cdots+(b_{N+1}-b_N) =b_{N+1}-b_1=(N+1)-1=N. \] The cancellation is exact, but the result still diverges because the remaining partial sums do not approach a finite number.
Check the original summand at \(n=1\): \(2!/1!-1!/0!=2-1=1\). This matches the simplified term. Notice also that the terms themselves do not tend to zero: every term is \(1\). That is consistent with divergence and illustrates why cancellation in a finite sum cannot replace checking the partial-sum limit.
A practical method and common mistakes
For a factorial series that may collapse, use the following sequence of steps. It keeps the algebra and the convergence question separate:
Expand a factorial such as \((n+k)!\) into its consecutive factors, then cancel common factorials only where they occur as factors.
Try a difference of adjacent reciprocal factorials or a difference of adjacent reciprocal integers. Put the proposed terms over a common denominator to verify the identity.
Expand enough terms to identify the cancellation, then write the exact first and last terms that remain.
Determine whether the simplified partial sums approach a finite number. If they do, that limit is the series sum; if not, the series diverges.
A common mistake is to conclude that a series converges as soon as its terms have been written as differences. For example, from the previous example one might reason, “The terms cancel, so the series converges.” That omits the uncancelled terms at the ends. In fact, its partial sum is \(b_{N+1}-b_1=N\), which does not approach a finite value. The correction is to write the finite sum explicitly and then take its limit.
Another error is to cancel factorial symbols when they are not common multiplicative factors. For instance, \((n+2)!\) is not just \(n!\); it is \((n+2)(n+1)n!\). Cancelling \(n!\) in \(n!/(n+2)!\) is valid only after writing that full product, and it leaves \(1/((n+1)(n+2))\), not \(1\). Expanding the factorial first makes the cancellation visible and prevents this mistake.
These methods matter because factorial notation can hide simple structure. A term that appears to involve rapidly growing quantities may reduce to a difference of small terms, and a long finite sum may then reduce to just a few endpoint terms. But only the partial-sum limit determines whether the infinite series has a finite sum.
Check Your Understanding
For each question, write out the relevant finite partial sum before deciding what happens as its upper index grows.
- Rewrite \(n!/(n+3)!\) as a product of consecutive factors, and then test whether it is the difference of two reciprocal products with adjacent endpoints.
- Verify algebraically that \((n+1)/(n+2)!=1/(n+1)!-1/(n+2)!\).
- Find the \(N\)th partial sum of \(\displaystyle\sum_{n=1}^{\infty}\left(1/n!-1/(n+1)!\right)\), and determine its limit.
- For a finite sum of the form \(\sum_{n=1}^{N}(b_n-b_{n+1})\), which terms remain after cancellation?
- Explain why writing terms as consecutive differences does not, by itself, prove that the corresponding infinite series converges.