Prove that $a_n=1+\frac{1}{1!} + \frac{1}{2!} +...+ \frac{1}{n!}$ converges using the Cauchy criterion

Question

Any tips on how to approach these kind of proof problems when a factorial is included?

Here is what I've tried,

By the Cauchy criterion the sequence converges if for every $\varepsilon>0$ there exists $N$ such that for all $n\ge N$ satisfies \begin{align}|a_{n+p}-a_n|&=\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\frac{1}{(n+3)!}+...+\frac{1}{(n+p)!}\\&<\frac{1}{n}+\frac{1}{(n+1)}+\frac{1}{(n+2)}+...+\frac{1}{(n+p-1)}\end{align} any ideas on how to continue from here if what i did is correct at all?

Answer 1

You can use this

HINT: $n!>2^n$.

More Hints:

$|a_n-a_m|=|\frac{1}{(m+1)!}+\frac{1}{(m+2)!}+\ldots+\frac{1}{(n)!}|$

$ \le |\frac{1}{(2^{m+1})}+\frac{1}{(2^{m+2})}+\ldots +\frac{1}{(2^{n})}|$

Now use the fact that $|r|^n\to 0 $ when $|r|<1$

Answer 2

Without Cauchy:

$a_n$ is clearly increasing, and as

$$n\cdot(n-1)\cdot(n-2)\cdots1\ge2^{n-1},$$

$$a_n=1+\frac{1}{1!} + \frac{1}{2!} +...+ \frac{1}{n!}\le1+\frac11+\frac12+\cdots\frac1{2^{n-1}}=1+\frac{1-\frac1{2^n}}{1-\frac12}<3.$$

Answer 3

Noting that $$ n!\ge (n-1)n $$ one has \begin{eqnarray} |a_{n+p}-a_n|&=&\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\cdots+\frac{1}{(n+p)!}\\ &\le&\frac{1}{n(n+1)}+\frac{1}{(n+1)(n+2)}+\cdots+\frac{1}{(n+p-1)(n+p)}\\ &=&\frac{1}{n}-\frac{1}{n+p}\\ &\le&\frac{1}{n}. \end{eqnarray} ...