How should I prove that the following sequence is cauchy?

Question

I have to prove from definition that the following sequence is Cauchy:

${1+\frac{1}{1!}+\frac{1}{2!}...+\frac{1}{n!}}$.

Definition of Cauchy sequence: A sequence $(a_n)$ is said to be a Cauchy sequence if given $\epsilon>0$, however small, there exists a $m\in N$ such that $|a_{m+p}-a_p|$<$\epsilon$ for all $n\geq m$ and for $p=1,2,3...$

Hint given in textbook is : $(n+1)!\geq 2^n$

So should I take $p=1?$ That would give me ${1+\frac{1}{1!}+\frac{1}{2!}...+\frac{1}{n!}+\frac{1}{(n+1)!}}$.

Then,

$|u_{n+1}-u_n|=\frac{1}{{(n+1)}!}\leq \frac{1}{2^n}$

Your definition has a typo! It should be $a_{m+p} - a_m$, not $a_p$. It makes a rather large difference to your subsequent calculation. — Erick Wong, Apr 24 '21 at 05:24

score 4 · Accepted Answer · answered Apr 24 '21 at 05:23

4

You need to prove it for any $p$, not just $p=1$, but you're on the right track.

$$|u_{n+p} - u_n| = \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \cdots + \frac{1}{(n+p)!} \le \frac{1}{2^n} + \frac{1}{2^{n+1}} + \cdots + \frac{1}{2^{n+p-1}} = \frac{1}{2^n} \frac{1-2^{-p}}{1-2^{-1}} \le \frac{1}{2^{n-1}}.$$

So, given $\epsilon$, can you choose $m$ such that $|u_{n+p} - u_n| < \epsilon$ for any $n \ge m$ and any $p$?

answered Apr 24 '21 at 05:23

angryavian

89,882

Oh! Thanks a lot. I get it now!! – Natasha J Apr 24 '21 at 05:26
Will $m$ be $ \frac{\log(epsilon)}{log\frac{1}{2}}+1$? – Natasha J Apr 24 '21 at 05:37
I know this is a dumb question but how did you get $2^{-p}?$ Shouldn't it be $2^{n+p}?$ – Natasha J Apr 24 '21 at 05:59
1

@NatashaJ I factored out a $\frac{1}{2^n}$ on the left, leaving $1 + \frac{1}{2} + \cdots + \frac{1}{2^{p-1}}$. – angryavian Apr 24 '21 at 06:07

How should I prove that the following sequence is cauchy?

1 Answers1