Let $x_n = \sum_{k=1}^{n} 1/k! = 1+1/2!+1/3!+...+1/n!$ for each $n \geq 0$. Show that the sequence $\left(x_n\right)$ is Cauchy.
This is what I have:
for $n>m$, \begin{align} &|x_n-x_m| \\ &=|(1+1/2!+...+1/m!+1/(m+1)!+...+1/(n-1)!+1/n!)-(1+1/2!+...+1/m!)| \\ &=1/n!+1/(n-1)!+...+1/(m+1)! \leq 1/2^{n-1} + 1/2^{n-2}+...+1/2^{m+1}+1/2^m \\ &= 1/2^m \cdot [1+1/2+...+1/2^{n-1-m}] \leq 1/2^{m-1} . \end{align} Therefore, for $\epsilon>0$, if $n$ is chosen and 1/2m<$\epsilon/2$<...<1/($2^{n-1}$)<$\epsilon$ so, |$x_n$-$x_m$|<$\epsilon$.
so, $x_n$ is Cauchy
Is this correct? if not can someone help me to make it correct.
The idea is correct. You should have stated explicitely that you are using that fact that $(\forall n\in\mathbb{N}):n!\geqslant2^{n-1}$. And, near the and, what you should choose is a natural $N$ such that $2^{-N}<\frac\varepsilon2$. It will follow then from your computations that$$n>m\geqslant N\implies\lvert x_n-x_m\rvert<\varepsilon.$$