Definition of null sequence I am using:

$(a_n) \to 0$ as $n \to \infty \iff$ given $\epsilon > 0$, there's $N$ such that $n > N \implies |a_n| < \epsilon.$

The sequences below are not null sequences because we can choose appropriate epsilons such that the absolute values of every term of a sequence after a certain point is greater or equal to epsilon.

a) $2, 1, 0, -0.1, 0.01, -0.001, 0.01, -0.001,\ldots, 0.01, -0.001,...\ldots$

b) $1, \frac12, 1, \frac14, 1, \frac18,\ldots$

The answers in my book(for proving the given sequences are not null) are $\epsilon = 0.01$ or less for (a) and $\epsilon = 1$ or less for (b).

I don't get these answers because in (a) if we choose $\epsilon = 0.01$, then there's always a term in the sequence such that $|-0.001| < \epsilon $ and in (b) if $\epsilon = 1$, then there's always a term $\frac 1n$ with $n \neq 0, 1$ such that $|\frac1n| < \epsilon$.

Please, explain why we choose these epsilons.

  • $\begingroup$ Instead of there's always a term you need all terms with $n\gt N$ to be within $\epsilon$ of $0$. Your problem is there is always a term with $|a_n| \not \lt \epsilon$ $\endgroup$ – Henry Mar 13 '16 at 22:00
  • $\begingroup$ This is how I am thinking: after we negate the definition of null sequence we need to find an $\epsilon \le |a_n| $ which holds for all $N$ after a $n$. If in (b), $\epsilon = 1$, then we can find such a $n$ which indexes a term, say, $\frac15$ which is smaller than $1$. Please, pinpoint exactly where I am tripping up s I can fix it. $\endgroup$ – user322231 Mar 13 '16 at 22:22

You're mixing up a "for all" and a "there exists." A sequence is null if for each $\epsilon$, there is some $N$ such that for all $n>N$ we have $\vert a_n\vert<\epsilon$.

So, for instance, looking at (b) and setting $\epsilon=1$: no matter what $N$ you pick, one or two terms later we have a "$1$", which is not $<\epsilon$.

Basically, think of it like this: a sequence is null if eventually every number in it is "small" - for every reasonable notion of "small." So these sequences aren't null because we keep having "large" (e.g. $\ge 1$ in (b)) terms.

  • $\begingroup$ Just to clarify: we do ignore the smaller numbers that occur after every $1$, right? For example, in (b), let $n = 5$. Then the next term is $\frac18$ which is smaller than $\epsilon = 1.$ And we ignore this because $1$ occurs over and over again in the sequence? $\endgroup$ – user322231 Mar 13 '16 at 22:51
  • $\begingroup$ @user322231 Yes - or rather, we don't ignore this fact, it's just not relevant. If we pick $\epsilon=1$, then for any $N$, we can find an $n>N$ such that $\vert a_n\vert\ge\epsilon$; this shows that $\epsilon=1$ is a counterexample to the definition of null-ness. The fact that small terms keep appearing doesn't matter; the point is that in a null sequence, large terms can't keep appearing. $\endgroup$ – Noah Schweber Mar 13 '16 at 22:54

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