Show that the sequence $1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}$ is Cauchy

Question

Show that the sequence $1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}$ is Cauchy. I'm not sure where to start with this problem, I know that if I can show that the sequence is convergent I can manipulate the inequality definition of the limit to have $|X_m-X_n|<\epsilon$ but I don't get how to prove the limit exists.

Duplicate of How to prove that $x_n=1+\frac{1}{2!}+\cdots+\frac{1}{n!}$ is a Cauchy sequence?. Also duplicates this question, whose accepted answer just offers a hint. — Brian M. Scott, Sep 23 '13 at 01:47
It converges by monotone convergence and boundedness, hence Cauchy. — Pedro, Sep 23 '13 at 01:50
@BrianM.Scott Sorry for posting a duplicate question. How can I remove my question? — Duiliath, Sep 23 '13 at 02:59

score 2 · Answer 1 · answered Sep 23 '13 at 01:49

2

Hint: Let $x_n=\displaystyle\sum_{k=1}^n\frac{1}{k!}$ and $m<n.$ Then $x_n-x_m=\displaystyle\sum_{k=m+1}^n\frac{1}{k!}\le\sum_{k=m+1}^n\frac{1}{2^k}$. This is a geometric sum.

answered Sep 23 '13 at 01:49

Show that the sequence $1+\frac{1}{2!}+\frac{1}{3!}+\cdots+\frac{1}{n!}$ is Cauchy

1 Answers1