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Let $(x_n)_{n\in\mathbb{N}}$ be a real sequence given by $$x_n=1+\frac{1}{2!}+\cdots+\frac{1}{n!}$$ I would like to prove that, for each $\varepsilon>0$, there exists $n_0\in\mathbb{N}$ such that if $n>m>n_0$ then $$|x_n-x_m|=\frac{1}{(m+1)!}+\cdots+\frac{1}{n!}<\varepsilon$$ In other words, I want to prove that $(x_n)$ is a Cauchy sequence. I know this sequence is a Cauchy sequence because it's convergent, but I want to prove it by definition of Cauchy sequence.

Thanks.

Pedro
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    You could estimate using the inequality ${1\over k!}\le {1\over 2^{k-1}}$. You should also know how to evaluate ${1\over 2^{m }}+{1\over 2^{m+1}}+\cdots+{1\over 2^{n-1}}$. – David Mitra Jul 30 '13 at 23:52

3 Answers3

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We have:

$$|x_n-x_m|=\frac{1}{(m+1)!}+\cdots+\frac{1}{n!}\leq \sum_{k=m}^\infty \frac{1}{2^k}=\frac{1}{2^{m-1}}$$ and $\frac{1}{2^{m-1}}$ can be made as small as we want.

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Let n and m be nonnegative integers (n,m≥1). We have: |x_(n+m)-x_n |=|(1+1/2!+⋯+1/n!+1/(n+1)!+⋯+1/(n+m)!)-(1+1/2!+⋯+1/n!)|

=|1/(n+1)!+⋯+1/(n+m)!|=1/(n+1)!+⋯+1/(n+m)!≤1/(n+1)^2 +⋯+1/(n+m)^2 =|y_(n+m)-y_n |

Since (y_n )n is a Cauchy sequence (1/k^2 ≤1/k(k-1) =1/(k-1)-1/k,k≥2), then there exists n_0∈IN such that n≥n_0: |y(n+m)-y_n |<ε Hence, we get: |x_(n+m)-x_n |<ε This achieves de proof.

Abdellatif Semmouri
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Hint: $\frac{1}{(m+1)!}+\ldots + \frac{1}{(m+k)!}< \frac{1}{m!}(\frac{1}{m}+\frac{1}{m^2}+\ldots+\frac{1}{m^k})$

MichalisN
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