The ratio test m I have to find out if $\sum_{n=1}^{\infty}{\frac{\log(n)}{n!}}$ is convergent.
I think it's convergent because (the ratio test):
$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}\frac{ \frac{log(n+1)}{(n+1)!} }{ \frac{log(n)}{n!}}=0<1$.
But how can I formally show it? I am looking for proof that the ratio does indeed have limit zero. I just use Maple to find it, but how can I show it?
 A: $\lim_{n \rightarrow \infty} (\log n)/n =0$.
There is a $n_0$ s.t. for $n \ge n_0$ :
$0< (\log n)/n<1/2$.
For $n >4:$
$(n-1)!>(n-1)(n-2) >$
$(n-n/2)(n-n/2)=n^2/4$.
For $n_1 >\max (4,n_0)$:
$\dfrac{\log n}{n!} < (1/2)(4/n^2)=(2/n^2)$.
The series converges (Comparison test).
A: To verify the ratio has limit zero:
For $ n > 1$,
$$
\begin{align}
\frac{a_{n+1}}{a_n} &= \frac{\log(n+1)}{(n+1)!} \cdot \frac{n!}{\log n} \\
&= \frac{\log(n(1+1/n)) }{(n+1)\log n} \\
&= \frac{\log n + \log (1+1/n)}{(n+1)\log n} \\
&= \frac{1}{n+1} + \frac{\log(1+1/n)}{(n+1)\log n}
\end{align}
$$
In this form it becomes clear each additive term on the right has limit zero.  
The first is elementary.  For the second term, the numerator has limit zero because the $\log$ function is continuous and in any case the denominator increases in size without limit as $ n\to \infty$.
I hope that is clear.
A: The ratio is 
$$\frac{\log (n+1)}{(n+1)!}\bigg/\frac{\log n}{n!}=\frac{\log (n+1)/\log n}{(n+1)!/n!}.$$
Now $(n+1)!/n!=n+1$, and we can crudely bound $1<\log(n+1)/\log n<\log (n^2)/\log n=2$ (for $n\geq2$). So the ratio is between $\frac{1}{n+1}$ and $\frac{2}{n+1}$ which both tend to $0$. By squeeze, so does the ratio itself.
