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shadsiei

Answered question

2021-08-20

Prove that following sequence is Cauchy, using just the definition
${\frac{(- 1)^{n}}{n}}_{n = 1}^{\infty}$

Answer & Explanation

Asma Vang

Skilled2021-08-21Added 93 answers

Consider the sequence
${a_{n}} = {\frac{(- 1)^{n}}{n}}_{n = 1}^{\infty}$
Let $ε > 0$
For n,m, $\in N$ if $n \geq m$ then $\frac{1}{n} \leq \frac{1}{m}$
Then
$| a_{n} - a_{m} | = | \frac{(- 1)^{n}}{n} - \frac{(- 1)^{m}}{m} |$
$\leq | \frac{(- 1)^{n}}{n} | + | \frac{(- 1)^{m}}{m} |$
$\leq \frac{1}{n} + \frac{1}{m}$
$\leq \frac{1}{n} + \frac{1}{m} = \frac{2}{m} < ε$ if $m > \frac{2}{ε}$
Let N be a positive integer such that $N \geq \frac{2}{ε}$
Then $| a_{n} - a_{m} | < ε$ for $= n \geq m$
So ${a_{n}} = {\frac{(- 1)^{n}}{n}}_{n = 1}^{\infty}$ is a Cauchy sequence

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