# Show that the sequence $(a_n)$ where $a_n = 1 + \frac{1}{2!} + \ldots + \frac{1}{n!}$ is of Cauchy.

Show that the sequence $(a_n)$ where $$a_n = 1 + \frac{1}{2!} + \ldots + \frac{1}{n!}$$ is of Cauchy.

Comments: I will use the definition $$\forall \epsilon > 0, \exists n_0 \in \mathbb{N}; n,m > n_0 \Rightarrow |a_n - a_m|< \epsilon.$$ If $n>m$ $$|a_n - a_m| = \left|\frac{1}{(n+1)!} + \ldots + \frac{1}{n!}\right| \leq \left|\frac{1}{(n+1)!}\right| + \ldots + \left|\frac{1}{n!}\right|$$ I do not know how to limit the $\epsilon$ and define $n_0$ without it depends of $n$ and $m$.

• You can't have the expression $|a_n-a_m|$ depend only on $n$. Check again your last displayed formula. Commented Sep 21, 2015 at 22:59
• Since you need to verify the definition for all $n,m>n_0$, it usually helps just letting the smaller of $m,n$ be $n_0$. Commented Sep 21, 2015 at 23:01
• Mario's answer is perfect, but here is another approach (different approaches are always good): Using the fact that $k!\geqslant2^k$ for each natural $k\geqslant4$ (you can prove this by induction on $k$) we have $$a_k\leqslant1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{2^4}+\cdots+\dfrac{1}{2^k}=1+\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{2^4}-\dfrac{1}{2^k}$$ whenever $k>3.$ Now let $\epsilon>0$ be given and let $N$ be such that $1/2^N<\epsilon/2.$ It follows that if $n,m\geqslant \max\{N,4\}$ then $\epsilon>1/2^m+1/2^n>|1/2^m-1/2^n|=|a_n-a_{m}|.$
– CIJ
Commented Sep 22, 2015 at 0:00

If $m>n>1$, then \begin{align} |a_m-a_n|&=\left|\frac{1}{(n+1)!}+\frac{1}{(n+2)!}+\ldots+\frac{1}{m!}\right|\\&<\left|\frac{1}{(n+1)!}+\frac{1}{2}\frac{1}{(n+1)!}+\ldots+\frac{1}{2^{m-n-1}}\frac{1}{(n+1)!}\right|\\ &=\frac{1}{(n+1)!}\sum_{k=0}^{m-n-1}\frac{1}{2^k}\\ &<\frac{1}{(n+1)!}\sum_{k=0}^{\infty}\frac{1}{2^k}\\ &=\frac{2}{(n+1)!}\\ &<\frac{1}{n!} \end{align} Then, given $\varepsilon>0$, by taking $n_0$ such that $n_0!>\max\left(\frac{1}{\varepsilon},1\right)$, we have $$m>n>n_0\quad\implies\quad|a_m-a_n|<\varepsilon$$