Show that given sequence is convergent. Show that given sequence$\{S_n\} $ is convergent. 
$$S_n = 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+.....+\dfrac{1}{(n-1)!}$$
is convergent.
My input:
$S_n = 1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+.....+\dfrac{1}{(n-1)!}<1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{2^2}...=1+1+1=3$
$S_n<3$
Given sequence is increasing and we have proved that it's bounded above by $3$. Thus making it a convergent sequence. Is it a correct approach to solve this problem? Any other way I could've solved this? 
 A: Yes, it is correct.
The standard way of proving that this sequence converges is to use the ratio test: if $(a_n)_{n\geqslant0}$ is a sequence of positive numbers such that $\lim_{n\to\infty}\frac{a_{n+1}}{a_n}<1$, then the sequence $a_0+a_1+a_2+\cdots$ converges.
A: Yes it is correct, as an alternative we have
$$\lim_{n\to \infty}S_n=\lim_{n\to \infty} \sum_{k=1}^n \frac 1{(k-1)!}=\sum_{k=1}^\infty \frac 1{(k-1)!}$$
which converges for example by comparison test with $\sum \frac 1{k^2}$.
A: Use the ratio test, so:
$$\frac{1}{(n+1)!} \div \frac{1}{n!}=\frac{n!}{(n+1)!}=\frac{1}{n+1}<1\forall n>0$$
A: Take
$$u_n=1+\sum_{i=1}^n\frac{1}{i!}$$
and
$$v_n=u_n+\frac{1}{nn!}$$
It is easy to prove that $(u_n)$ and $(v_n)$ are adjacent and therefore convergent.
You can also use this to prove that the limit $e\notin \Bbb Q$.
A: An option:
Let $k\ge 2$.
$\dfrac{1}{k!} \le \dfrac{1}{k(k-1)} = \dfrac{1}{k-1} - \dfrac{1}{k}.$
We have
$\displaystyle {\sum_{k=2}^{n}} \dfrac{1}{k!} \le \displaystyle{ \sum_{k=2}^{n}}[ \dfrac{1}{k-1}-\dfrac{1}{k}] =$
$1- \dfrac{1}{n}.$
1) $S_n=\displaystyle{ \sum_{k=0}^{n-1}}\dfrac{1}{k!}$ is strictly increasing,
2) $S_n$ is bounded above by $3$ (why?),
hence convergent.
