Question: Prove the convergence of sequence using Cauchy general principle of convergence. $$\frac{1}{1} + \frac{1}{2!} + \frac{1}{3!} + . . + \frac{1}{n!}.$$
My attempt:
$$ | s_2 - s_1 | = \frac{1}{2!} \\ | s_3 - s_2 | = \frac{1}{3!}\;\;\;\;\text{i.e.}\;\;\;\;| s_2 - s_1 | \leq| s_2 - s_1 | \\ | s_4 - s_3 | = \frac{1}{4!}\;\;\;\;\text{i.e.}\;\;\;\;| s_4 - s_3 | \leq | s_2 - s_1 | \\ \text{Finally,}\;\;\;\;| s_{n+1} - s_n | \leq | s_2 - s_1| $$ Now for $\epsilon>0$, there exists $n\geq m$ such that $$ | s_n - s_m | = | (s_n - s_{n-1}) + ( s_{n-1} - s_{n-2}) + \cdots + (s_{m+1} - s_m) |$$ or $$ \leq |s_n - s_{n-1}| + |s_{n-1} - s_{n-2}| + \cdots + |s_{m+1} - s_m|$$or $$ \leq |s_n - s_{n-1}| + |s_{n-1} - s_{n-2}| + \cdots + |s_{m+1} - s_m|$$ or $$ \leq |s_2 - s_1| + |s_2 - s_1| + \cdots + |s_2 - s_1|$$or $$ \leq \frac{n-m}{2}.$$ Challenge: From here on I am unable to proceed and always getting stuck in assuming values for $\epsilon$ and $n$. This is happening for similar questions, too. Please help me to the correct frame of thinking for such questions, and also help in completing above question. Thank you.
If $m>n$, then\begin{align}s_m-s_n&=\frac1{(n+1)!}+\frac1{(n+2)!}+\cdots+\frac1{m!}\\&=\frac1{(n+1)!}\left(1+\frac1{n+2}+\frac1{(n+2)(n+3)}+\cdots+\frac1{(n+2)(n+3)\cdots m}\right)\\&\leqslant\frac1{(n+1)!}\left(1+\frac1{n+2}+\frac1{(n+2)^2}+\cdots\right)\\&=\frac1{(n+1)!}\cdot\frac1{1-\frac1{n+2}}\\&=\frac{n+2}{(n+1)\times(n+1)!}.\end{align}Now, take $\varepsilon>0$. Since$$\lim_{n\to\infty}\frac{n+2}{(n+1)\times(n+1)!}=0,$$ there is some $N\in\Bbb N$ such that$$n\geqslant N\implies\frac{n+2}{(n+1)\times(n+1)!}<\varepsilon.$$So, if $m,n\geqslant N$, you hvae three possibilities: