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?
 A: 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 $e^3 - 1$).
A: Since $\sum_{n=0}^{m}\frac{3^n}{n!}$ converges to $e^3$ as $m \rightarrow \infty$,
$\sum_{n=0}^{m}\frac{3^n}{n!}-1$ converges to $e^3-1$ as $m \rightarrow \infty$
That is $\sum_{n=1}^{m}\frac{3^n}{n!}$ converges to $e^3-1$ as $m \rightarrow \infty$
A: When saying that you have 'reduced it' to $e^3 -1$ - which none the less is correct and thus the series converges to that particular value - you most likely used that $e^x = \sum_n x^n/n!$, which luckily converges everywhere, so you are allowed to evaluate it for $x = 3$; Unlike the series $\sum_n x^n$ which cannot be evaluated for e.g. $x = 3$. But by just stating that the series expansion for $e^x$ converges for every $x$, you have already solved the problem!
Since you ask of convergence, to find the convergence radius, use the ratio test as suggested;
$$\lim_{n\to \infty} \left \lvert \frac{x^{n+1}/(n+1)!}{x^n/n!}\right \rvert = \lim_{n\to \infty} \frac{x}{n+1} = 0 < 1, \quad \forall x\in \Bbb{R},
$$
so in particular, the series for $e^x$ converges for $x = 3$.
A: If for all n, n not equals 0, then the following rules apply:
Let L = lim (n -- > inf) | an+1 / an |.
If L < 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.
