Test convergence of $\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n!}$

Expert intro: Factorials grow so fast that they usually dominate any power of $x$. That is the clue here. The alternating sign does not make the convergence harder; the factorial already does the heavy lifting.

Detailed walkthrough

We have

$\sum_{n=1}^{\infty} \frac{(-1)^n x^n}{n!} = \sum_{n=1}^{\infty} \frac{(-x)^n}{n!}.$

Step 1: Compare with the exponential series

Recall that

$e^t=\sum_{n=0}^{\infty} \frac{t^n}{n!}.$

If we take $t=-x$, then

$e^{-x}=\sum_{n=0}^{\infty} \frac{(-x)^n}{n!}.$

Our series starts at $n=1$, so it is just missing the $n=0$ term, which is 1:

$\sum_{n=1}^{\infty} \frac{(-x)^n}{n!}=e^{-x}-1.$

Step 2: Convergence

Because the exponential series converges for every real $x$, this series also converges for every real $x$.

So the radius of convergence is

$\boxed{R=\infty}.$

Final answer

• Interval of convergence: $(-\infty,\infty)$
• Sum: $e^{-x}-1$

💡 Pitfall guide

A small but common mistake is to think the alternating sign is what makes it converge. Here that is not the main reason. The factorial in the denominator is enough to force convergence for all $x$. Another easy slip is forgetting the missing $n=0$ term when identifying the exponential series.

🔄 Real-world variant

If the series started at $n=0$, then the sum would be exactly $e^{-x}$. If the factorial were replaced by a polynomial like $n^2$, the convergence would be much more delicate and would depend on $x$.

🔍 Related terms

exponential series, factorial growth, radius of convergence