How to Test Whether the Series \(\sum \frac{1}{n+n!}\) Converges

Key concept: This is a nice comparison-test problem. The factorial in the denominator grows so fast that the terms become much smaller than a familiar p-series term.

Step by step

Step 1: Compare the term with something simpler

Let

$a_n=\frac{1}{n+n!}.$

Because $n+n!\ge n!$, we have

$0<\frac{1}{n+n!}\le \frac{1}{n!}.$

Step 2: Use a known convergent series

The series

$\sum_{n=1}^{\infty}\frac{1}{n!}$

converges.

Since $a_n$ is positive and bounded above by a convergent series term-by-term, the comparison test gives

$\sum_{n=1}^{\infty}\frac{1}{n+n!} \text{ converges.}$

A quick ratio-test check

If you try the ratio test,

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

and the factorial growth drives this ratio toward 0, which is also consistent with convergence.

Pitfall alert

A common slip is to compare $\frac{1}{n+n!}$ with $\frac{1}{n}$. That is too weak and can distract you from the fact that $n!$ dominates everything here. The clean move is to compare directly with $\frac{1}{n!}$ or use the ratio test.

Try different conditions

If the denominator were changed to $n+(n!)^2$, the same idea still works: the terms are even smaller, so comparison with $\frac{1}{n!}$ or $\frac{1}{(n!)^2}$ would again show convergence. If the denominator were only $n^2+n$, then you would need a different test, because the factorial advantage would be gone.

Further reading

comparison test, factorial growth, absolute convergence