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?