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The sequence: $-2, 0, -1, 0, -1/2, 0, -1/4, 0, -1/8, 0, -1/16, 0\ldots$

Heyy. So I was a little confused about this question. How would this have an existing limit if it keeps alternating between a value and $0$? I know if you have a sequence of $1, -1, 1, -1\ldots$ that would not have a limit.

So why is the sequence above have a limit?

Thank you in advance for any help!

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    $\begingroup$ It has a limit if the alternate sequence also converges to 0. $\endgroup$
    – Doug M
    Commented Feb 5, 2021 at 22:47
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    $\begingroup$ Have you tried to write down the sequence like $a_n=\ldots$ ? $\endgroup$
    – callculus42
    Commented Feb 5, 2021 at 22:48
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    $\begingroup$ Look again at the $\epsilon$-$N$ definition of limit... "A sequence is said to converge to a limit $L$ iff for every $\epsilon>0$ there exists an $N$ such that if $n>N$ then $|a_n-L|<\epsilon$. That is to say... however close we want to talk about there will always be some point in the sequence such that for that point on we will always be at least that close to the limit. $\endgroup$
    – JMoravitz
    Commented Feb 5, 2021 at 22:58
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    $\begingroup$ Here, if we want to talk about being at least within $\frac{1}{2}$ of zero, that starts happening and continues to happen from then on by the sixth term in the sequence. If we want to talk about being at least $\frac{1}{100}$ that happens by the eighteenth term and from then on, and so on so forth. If I were to ask you when it starts being within $\frac{1}{10^{10}}$ of zero from some point on, you could feasibly come up with that value too, even if not getting it exactly. It is for this reason we say it converges to zero. $\endgroup$
    – JMoravitz
    Commented Feb 5, 2021 at 23:01

2 Answers 2

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Just remember the definition of a sequence limit in term of $\varepsilon$-$\delta$: $(a_n)_{n} \subset \mathbb R$ converges to $a \in \mathbb R$ if and only if for all $\varepsilon > 0$, there exists $N_\varepsilon \in \mathbb N$ such that for all $n \ge N_\varepsilon$ we have $$|a - a_n| < \varepsilon.$$ In this case, $a = 0$ and $$ a_n = \begin{cases} -1/2^{n/2} & \text{if } n \text{ is even},\\ 0 & \text{otherwise}. \end{cases} $$ Let $\varepsilon > 0$. For $N_\varepsilon = \lceil 2\log_2(1/\varepsilon)\rceil$, we have, $\forall n \ge N_\varepsilon$, that $$\left|a - a_n\right| = \left|a_n\right| < \varepsilon.$$

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$\left<-2, 0, -1, 0, -1/2, 0, -1/4, 0, -1/8, 0, -1/16, 0\ldots\right>$

How would this sequence have an existing limit if it keeps alternating between a value and $\mathbf0$?

I know that the sequence $\left<1, -1, 1, -1\ldots\right>$ does not have a limit.

To be clear, $\left<-2, 0, -1, 0, -1/2, 0, -1/4, 0, -1/8, 0, -1/16, 0\ldots\right>$ is not alternating between some (constant) value and $\mathbf0,$ unlike $\left<1, -1, 1, -1\ldots\right>,$ which is alternating between two constant values. This crucial difference between them means that the latter cannot have a limit, while the former may.

Rather, the first sequence is approaching $0$ from the left; this means that its limit is zero (regardless of whether it ever reaches $0$).

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