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Let ${s_n}$ be a sequence such that ${s_n}=1+\frac{1}{2!}+...+\frac{1}{n!}$

Prove that ${s_n}$ is a Cauchy sequence.

My work thus far:

$|s_m-s_n|$=$|1+\frac{1}{2!}+...+\frac{1}{n!}+...+\frac{1}{(n+k)!}-1-\frac{1}{2!}+...+\frac{1}{n!}|<\epsilon$

$=\frac{1}{(n+k)!}+...$

Now, this this possible(?):

Given $\epsilon>0$ and letting $m,n>n_0, m=n+k$ I'm going to a choose $n_0$ such that $\frac{1}{(n+k)!}<\epsilon$, so I'm choosing $n_0>\frac{1}{\epsilon}$ assuming that $n_0=m+n$

sillyme
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1 Answers1

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If your intention is that $m=n+k$, you need to say so; as it is, you never defined $k$, so the expressions involving $k$ are meaningless. And if $m=n+k$, your calculation is incorrect:

$$|s_m-s_n|=s_m-s_n=\frac1{(n+1)!}+\frac1{(n+2)!}+\ldots+\frac1{(n+k)!}\;,$$

not just the last term of that sum.

HINT: $0<s_{n+1}\le\frac12s_n$ for all $n\ge 1$, and it’s easy to show that a geometric sequence with ratio $\frac12$ is Cauchy.

Brian M. Scott
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