Determine whether the given series is convergent. I have a series:
$$ \sum_{n=1}^\infty \frac{3^n}{n!}$$
The task is to investigate if this series converges or diverges. I know that if $\lim_{n\to\infty}\frac{3^n}{n!}$ is infinity or a non-real number, then it diverges, otherwise it converges.
I can simply look at the expression and conclude that the sum of $1*2*3*4*5*6*...*n$ will grow faster than $ 3*3*3*3*3*...$ And therefore I know that the expression $\frac{3^n}{n!}$ is moving towards 0, the serial will therefore be convergent. 
I am not shure that this calculation is valid enough. Do I have to calculate this in some other way or is this answer enough?
 A: Let me make some points:


*

*For a series $\sum{a_n}$ to converge(or exist) then $\lim\limits_{n\rightarrow\infty}{a_n}=0$. And if the value of the limit is any other real number other that zero or the limit does not exist the series diverges. Also this does not mean that $\lim\limits_{n\rightarrow \infty}{a_n}=0$ implies that the series converges, for ex: consider the series $\sum\limits_{n=1}^{\infty}{\frac1n}$.

*You are right in your analysis that the denominator of the fraction $\frac{3^n}{n!}$ grows faster than the denominator and hence $\lim\limits_{n\rightarrow \infty}{a_n}=0$, but as you can see from my previous point this does not imply that the sum of the sequence converges.

As I think you might are more familiar with convergence of sequences( as you predicted that $\frac{3^n}{n!}$ converges). Let us see how we can analyze this series as a sequence.
Consider a sequence $p_n=\sum\limits_{k=1}^{n}{\frac{3^n}{n!}}$. Here $p_n$ is also known as partial sum of the series, it is easy to prove that $p_n$ converges if and only if $\sum{\frac{3^n}{n!}}$ converges.
Denoting $a_n=\frac{3^n}{n!}$, note that $\lim\limits_{n\rightarrow \infty}{\frac{a_{n+1}}{a_n}}=0$. 
So what does that mean?. We can say that for every $1>\varepsilon>0$(Although for every $\varepsilon>0$ there should exist a $\delta$, I am additionally imposing the condition $\varepsilon <1$ for reasons you will see in a moment) there exists an $N$ such that for $n>N$ , $a_{n+1}<\varepsilon \cdot a_n$. 
Now as terms are finite let $k=\sum\limits_{k=1}^{N}{a_n}$ for  $n>N$, $$p_n=k+a_N\left(1+\varepsilon+\varepsilon^2+\cdots+\varepsilon^{n-N}\right)<k+a_N\left(1+\varepsilon+\varepsilon^2+ \cdots  \right)=k+\frac{a_N}{1-\varepsilon}$$
Now this proves that $p_n$ is bounded(bounded by $k+\frac{a_N}{1-\varepsilon}$), now $p_n$ also has an interesting property that $p_n$ is monotonically increasing which means that $p_{n+1}>p_n$ (as every term of the series is positive). So as you might be expecting a bounded monotonic sequence converges (see here).
Hence $p_n$ convergence implies the series converges.

In the previous section, I wrote a descriptive proof although this can be done using something known as ratio-test, which unfortunately you might be unfamiliar with, so I have illustrated a specific proof of the ratio-test here, if you are following it I recommend you study ratio test and the proof must be looking easier now.
A: If you look at the power expansion of exponential function (http://en.wikipedia.org/wiki/Exponential_function), i.e., 
$$e^x=\sum_{n=0}^{\infty}\frac{x^n}{n!}$$
You notice a big "resemblance". Yes, in fact
$$e^3=\sum_{n=0}^\infty\frac{3^n}{n!}=1+\sum_{n=1}^\infty\frac{3^n}{n!}$$
Clearly then $$\sum_{n=1}^\infty\frac{3^n}{n!}=e^3-1.$$
A: The ratio test shows that 
$$
\frac{3^{n+1} / (n+1)^{n+1}}{3^n/n^n}=3\frac{n!}{(n+1)!} <1 \text{ for $n>2$}
$$
 hence it converges... to a interesting value.
