For which values of x does the series sum n!x^n converge?

Key concept: This series is one of those cases where the factorial grows so fast that the ratio test gives a very clean answer. The key is not to overcomplicate it: check what happens as n gets large.

Step by step

Consider the series

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

Let

$a_n=n!x^n$

Apply the ratio test:

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

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

• If $x\neq 0$, then $|(n+1)x|\to\infty$
• So the series diverges for every nonzero $x$

If $x=0$, the series becomes

$\sum_{n=0}^{\infty} n!\cdot 0^n$

which is convergent, since all terms after the first are zero.

Final answer

The series converges only when

$x=0$

Pitfall alert

A frequent slip is to say the ratio test is inconclusive because the limit is infinite. It is not inconclusive here: a ratio limit greater than 1, including infinity, means divergence. Also, remember to check the special case $x=0$ separately.

Try different conditions

If the series were instead $\sum_{n=0}^{\infty} \frac{x^n}{n!}$, the conclusion would be completely different: that series converges for every real $x$. The factorial in the denominator helps convergence, while the factorial in the numerator forces divergence unless $x=0$.

Further reading

ratio test, factorial series, power series