Prove that the sequence $x_n=1+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{n!}$ is cauchy?
Here is my attempt and I used this How to prove that $x_n=1+\frac{1}{2!}+\cdots+\frac{1}{n!}$ is a Cauchy sequence? and complete my proof.
Let $\varepsilon>0$ and $n>m>N(\varepsilon)>\max\{4,2+log_2(\frac{1}{\varepsilon})\}$. Consider,
$$ |x_n-x_m|=\bigg|\frac{1}{(m+1)!}+\frac{1}{(m+2)!}+...+\frac{1}{n!} \bigg|$$
Since $2^n<n!$ for $n\geq 4$,
$$ |x_n-x_m|=\bigg|\frac{1}{(m+1)!}+\frac{1}{(m+2)!}+...+\frac{1}{n!} \bigg|<\frac{1}{2^{m+1}}+\frac{1}{2^{m+2}}+...\frac{1}{2^{n}}$$
$$ |x_n-x_m|=\bigg|\frac{1}{(m+1)!}+\frac{1}{(m+2)!}+...+\frac{1}{n!} \bigg|<\frac{1}{2^{m+1}}+\frac{1}{2^{m+2}}+...\frac{1}{2^{n}}<\sum_{k>m}\frac{1}{2^k}=\frac{1}{2^{m-2}}<\varepsilon$$
whenever $m>N(\varepsilon)>\max\{4,2+log_2(\frac{1}{\varepsilon})\}$
That is for all $\varepsilon>0$ there exists $N(\varepsilon) \in N$ such that $|x_n-x_m| <\varepsilon$ whenever $n>m>N(\varepsilon)>\max\{4,2+log_2(\frac{1}{\varepsilon})\}$
Can anyone verify my attempt?