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## Answered question

2021-08-20

Prove that sequence is Cauchy. Use defenition only
${\left\{\frac{n}{n+1}\right\}}_{n=1}^{\mathrm{\infty }}$

### Answer & Explanation

Nichole Watt

Skilled2021-08-21Added 100 answers

$\left\{{a}_{n}\right\}={\left\{\frac{n}{n+1}\right\}}_{n=1}^{\mathrm{\infty }}$
Let $\epsilon >0$
For n,m, $\in \mathbb{N}$ let $n\ge m$
Then
$|{a}_{n}-{a}_{m}|=|\frac{n}{n+1}-\frac{m}{m+1}|$
$=|\frac{n\left(m+1\right)-m\left(n+1\right)}{\left(n+1\right)\left(m+1\right)}|$
$=|\frac{n-m}{\left(n+1\right)\left(m+1\right)}|$
Since, $nm>\left(n+1\right)\left(m+1\right)$ and n−m $|{a}_{n}-{a}_{m}|<|\frac{n}{nm}|<\frac{1}{m}<\epsilon$
if $m>\frac{1}{\epsilon }$
Let, N be a positive integer, such that, $N\ge \frac{1}{\epsilon }$
Then, P$|{a}_{n}-{a}_{m}|<\epsilon$, for $n\ge m$.
So $\left\{{a}_{n}\right\}={\left\{\frac{n}{n+1}\right\}}_{n=1}^{\mathrm{\infty }}$ is a Cauchy sequence

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