# Is the sequence $(1 + (-1)^n)$ Cauchy?

We're supposed to prove or disprove the statement that $(1 + (-1)^n)$ is a Cauchy sequence. I don't think it is because this function oscillates and doesn't converge, so I don't think it's Cauchy. Why or why not is this a Cauchy sequence? Thanks.

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Hint: Cauchy $\iff$ convergent, for sequences of real numbers –  Jean-Claude Arbaut Dec 7 '13 at 0:32
How could I show that this sequence is not convergent? –  anonymous Dec 7 '13 at 0:35
It has two subsequences that converge to different values (actually they are constant). Try with terms with odd and even indices. –  Jean-Claude Arbaut Dec 7 '13 at 0:36

Note that $\left \vert a_{n+1} - a_n \right \vert = 2$ for all $n$. So is this sequence Cauchy?

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You can apply the definition of Cauchy sequence directly. Note that if $a_n=1+(-1)^n$, then

$$a_n=\begin{cases} 2,&\text{if }n\text{ is even}\\ 0,&\text{if }n\text{ is odd}\;. \end{cases}$$

Let $\epsilon>0$. If the sequence is Cauchy, there must be an $m\in\Bbb N$ such that $|a_k-a_\ell|<\epsilon$ whenever $k,\ell\ge m$. Now consider $|a_k-a_{k+1}|$ for any $k\in\Bbb N$; it’s equal to ... ?

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If the series converges to a limit $L$, then for sufficiently big values of $n$, you'd have $(1+(-1))^n$ between $L\pm1/10$. Thus two consecutive terms would differ from each other by less than $2/10$. But $0$ and $2$ do not differ by less than $2/10$. Hence the sequence diverges. Therefore it's not a Cauchy sequence.