Convergence of series $\sum\limits_{n=0}^\infty\frac{3n}{n!}$ 
Investigate the series for convergence and if possible, determine its
  limit: $\sum\limits_{n=0}^\infty\frac{3n}{n!}$

My solution
$$\sum\limits_{n=0}^\infty\frac{3n}{n!} = \sum\limits_{n=0}^\infty\frac{3n}{(n-1)!n} = \sum\limits_{n=0}^\infty\frac{3}{(n-1)!}$$
But since $n \to \infty$, the sum is the same as
$$\sum\limits_{n=0}^\infty\frac{3}{n!} = 3\sum\limits_{n=0}^\infty\frac{1}{n!} = 3\sum\limits_{n=0}^\infty\frac{1}{n!}1^n$$
which is the definition of $e$, so the series is convergent and it equals to $3e^1$.
Is this proof correct? How could I improve it by making it more formal?
 A: If you replace $$\sum_{n=0}^\infty \frac{3n}{n!}=\sum_{n=0}^\infty\frac{3n}{(n−1)!n}$$ 
by
$$\sum_{n=0}^\infty \frac{3n}{n!}=\sum_{n=1}^\infty \frac{3n}{n!}=\sum_{n=1}^\infty\frac{3n}{(n−1)!n},$$ 
you can just replace the vague $n \to \infty$ line by an equal sign.
A: This should be a comment, but it is a little long.
Note that the term corresponding to $n=0$ is $0$. This is because $0!$, by definition, is equal to $1$. So, as pointed out by Phira, our sum is equal to $\sum_{n=1}^\infty \frac{3n}{n!}$.  As in your answer, this simplifies to $\sum_{n=1}^\infty \frac{3}{n-1}!$. Writing $k$ for $n-1$, we see that our sum is $\sum_{k=0}^\infty \frac{3}{k!}$, which we recognize. 
Conceivably, this is not quite the desired answer! Perhaps the problem setter expects us to investigate the series for convergence while pretending we don't know the sum, and as a second part, to find the sum. If so, the Ratio Test quickly tells us that the series converges.
Even though you end up with the right number, the first line of your answer is problematic. For in the case $n=0$, you are cancelling $0$'s (in general forbidden). Then your first term is $\frac{3}{(-1)!}$, whatever that may mean. Then for no clear reason the $(n-1)!$ gets replaced by $n!$. 
A: Beside the point that @Phira noted: $$\sum_{n=0}^\infty\frac{3n}{(n−1)!n}\to\sum_{n=1}^\infty\frac{3n}{(n−1)!n}$$ I think, since the partial sums of the latter series is the same patial sums for $\exp(1)$, then your claim looks right.
A: $\sum_{n=0}^\infty \frac{3n}{n!} < \sum_{n=0}^\infty \frac{3^n}{n!}=e^3$
