Find the interval of convergence for the power series \(\sum \frac{x^n}{n!}\)

Expert intro: A fast way to handle this series is to compare it with the exponential function. The factorial in the denominator usually makes convergence very generous, but it still helps to check it carefully with a standard test.

Detailed walkthrough

We want the interval of convergence of

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

Step 1: Use the Ratio Test

Let

$a_n=\frac{x^n}{n!}.$

Then

$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{x^{n+1}}{(n+1)!}\cdot\frac{n!}{x^n}\right|=\frac{|x|}{n+1}.$

Now take the limit as $n\to\infty$:

$\lim_{n\to\infty}\frac{|x|}{n+1}=0.$

Since $0<1$ for every real value of $x$, the series converges absolutely for all $x$.

Step 2: State the interval

Because convergence holds for every real number,

$(-\infty,\infty)$

is the interval of convergence.

Final answer

The series converges for all real $x$, so the interval of convergence is

$(-\infty,\infty).$

💡 Pitfall guide

A common mistake is to treat this like a geometric series and look for a boundary value such as $|x|<1$. That pattern does not apply here because the factorial grows much faster than any power of $x$.

Another slip is forgetting that the Ratio Test gives absolute convergence for every fixed $x$, including large positive or negative values.

🔄 Real-world variant

If the series were instead

$\sum_{n=0}^{\infty} \frac{(ax)^n}{n!},$

the same Ratio Test still gives

$\left|\frac{a_{n+1}}{a_n}\right|=\frac{|ax|}{n+1}\to 0,$

so the interval of convergence would still be all real numbers. The parameter $a$ changes the terms, but not the final convergence result.

🔍 Related terms

ratio test, absolute convergence, power series