What Is the Sum of the Alternating Reciprocal-Factorial Series?

The phrase “alternating series of reciprocals of factorials” can refer to two closely related series, depending on the starting index and first sign. For the common version that begins $1/1!$, the sum is $1-1/e$. The version that begins with the extra term $1/0!=1$ instead sums to $1/e$.

The key is to match the terms carefully with the Taylor series for the exponential function.

Use the exponential series

The Maclaurin series for $e^x$ is

$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!} =1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$$

Substituting $x=-1$ gives

$$e^{-1}=\sum_{n=0}^{\infty}\frac{(-1)^n}{n!} =1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots.$$

Since $e^{-1}=1/e$, this identifies the sum when the series starts at $n=0$. The $n=0$ term is $1/0!=1$, so removing that term changes the sum by 1.

The series starting with $1/1!$

A common interpretation of the question is

$$S=\frac{1}{1!}-\frac{1}{2!}+\frac{1}{3!}-\frac{1}{4!}+\cdots.$$

The exponential expansion above has the opposite sign on every term after the initial 1. In fact,

$$\frac{1}{e}=1-S.$$

Therefore,

$$\boxed{S=1-\frac{1}{e}}.$$

Numerically, $1-1/e\approx 0.63212056$.

More generally, for an integer $N\geq 1$, the first $N$ terms are

$$S_N=\sum_{n=1}^{N}\frac{(-1)^{n+1}}{n!}.$$

The infinite sum follows directly from the exponential series; no special factorial identity is needed.

A worked partial-sum example

Take the first four terms:

$$S_4=1-\frac{1}{2}+\frac{1}{6}-\frac{1}{24} =\frac{24-12+4-1}{24} =\frac{15}{24}=\frac{5}{8}=0.625.$$

The exact sum is about $0.63212$, so this four-term estimate is low by about $0.00712$. The terms decrease in size, and the signs alternate. The alternating-series error bound says that the error is no greater than the first omitted term:

$$|S-S_4|\leq \frac{1}{5!}=\frac{1}{120}\approx 0.00833.$$

The observed error is within that bound. Adding the fifth term gives $S_5=5/8+1/120=19/30\approx 0.63333$; this estimate is high, and its error is at most the next term, $1/6!$.

If the series starts at $n=0$

If the intended series is

$$1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\cdots,$$

then it is exactly the Maclaurin expansion of $e^{-1}$. Its sum is

$$\boxed{\frac{1}{e}}.$$

A different sign convention can also occur. For example, $-1/1!+1/2!-1/3!+\cdots$ is the negative of the series starting with $1/1!$, so its sum is $1/e-1$.

Why the series converges

The absolute values of the terms are $1/n!$. For the series starting at $n=1$, their sum is $e-1$, by the expansion of $e^1$; for the version starting at $n=0$, it is $e$. Thus both series converge absolutely. Alternatively, the ratio of consecutive absolute terms is

$$\frac{1/(n+1)!}{1/n!}=\frac{1}{n+1},$$

which tends to zero. Thus the terms decrease to zero, and the alternating-series test also guarantees convergence. Absolute convergence means that changing the order of the terms would not change the sum, although the displayed alternating order is useful for estimating it.

Common mistakes

• Confusing the two starting indices. The series from $n=0$ includes $1/0!=1$ and sums to $1/e$. The positive-first series from $n=1$ omits that term and sums to $1-1/e$.
• Dropping the leading 1 incorrectly. Removing the $n=0$ term from the expansion of $e^{-1}$ leaves a series whose terms have the opposite signs from the positive-first version; hence the relation is $1/e=1-S$.
• Assuming a partial sum is exact. A finite alternating sum is an approximation. Its error is bounded by the first omitted term, not generally zero.
• Treating the factorial as a power. The denominator is $n!=1\cdot2\cdots n$, so terms shrink faster than a geometric sequence with a fixed ratio.

When evaluating a particular version, write its first two or three terms and compare them with the expansion of $e^{-1}$. That makes the indexing and sign adjustment explicit.