Does the sequence $\frac{3^n}{n!}$ converge or diverge? Consider the sequence $\frac{3^n}{n!}$ where $n$ varies from $1$ to $\infty$. Does the corresponding  series converge or diverge?


  $$
\sum_{n=1}^\infty \frac{3^n}{n!}.
$$


I reduced this to $(e^3 - 1)$. But how should I decide if this converges/diverges?
Good news!

You have already solved the problem.

As you correctly note, this is $e^3 - 1$.

That's because the series for $e^x$ converges everywhere! So you are done, it converges (specifically to $e^3 - 1$).
If for all n, n not equals 0, then the following rules apply:
Let L = lim (n -- > inf) | an+1 / an |.
If L &lt; 1, then the series sum (1..inf) an converges.
If L > 1, then the series sum (1..inf) an diverges.
If L = 1, then the test in inconclusive.
Here the limit lim (n -- > inf) | an+1 / an |. Is Zero. hence the series is converging.
This is the ratio test.