Prove that sequence is Cauchy. Use defenition only {nn+1}n=1∞

{an}={nn+1}n=1∞ Let ε&gt;0 For n,m, ∈N let n≥m Then |an−am|=|nn+1−mm+1| =|n(m+1)−m(n+1)(n+1)(m+1)| =|n−m(n+1)(m+1)| Since, nm&gt;(n+1)(m+1) and n−m |an−am|&lt;|nnm|&lt;1m&lt;ε if m&gt;1ε Let, N be a positive integer, such that, N≥1ε Then, P|an−am|&lt;ε, for n≥m. So {an}={nn+1}n=1∞ is a Cauchy sequence

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Algebra
Algebra I
Sequences

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Answered question

2021-08-20

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

Answer & Explanation

Nichole Watt

Skilled2021-08-21Added 100 answers

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

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