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The question is from my homework. It is to find the limit of the sequence as n approaches infinity:

$$a_n = \frac{(-1)^n}{3\sqrt{n}}$$

Please help! I have no idea how to approach this question. Thanks

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Why don't you just try to plug in some (large) values of $n$ to see what happens? – Hans Lundmark Oct 28 '12 at 15:20
up vote 0 down vote accepted

Hint: your numerator, for every $n$, is bounded by the interval $[-1,1]$, while your denominator tends to infinity...

More rigorously:

Theorem. Let's assume that sequence $a_{n}$ has $k$ subsequences satisfying the following properties:

  1. All of the subsequences converge to $g$;
  2. Every term of $a_{n}$ belongs to one of those subsequences

Then $a_{n}$ also converges to $g$.

Application: let's choose two subsequences: $a_{2n}$ and $a_{2n-1}$. We have:

$$a_{2n}=\frac{1}{3\sqrt{n}}\qquad a_{2n-1}=\frac{-1}{3\sqrt{n}}$$

We can easily notice that:

  1. Every term of the sequence $a_{n}$ belongs to one of the above subsequences, and
  2. Both $\lim_{n\to\infty}a_{2n}=0$ and $\lim_{n\to\infty}a_{2n-1}=0$

Hence, the limit of $a_{n}$ is also $0$.

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so isn't this limit zero? – Jaden Q Oct 28 '12 at 8:45
@JadenQ - yes, it is. – Johnny Westerling Oct 28 '12 at 8:50

While you would notice that numerator is bounded and denominator is not, hence $a_n$ could have only bounded value, this is not sufficient to have a limit. For having a limit, it should also be converging to some point and not bouncing back and forth between finite points, say +2 and -2. Luckily, +0 and -0 are the same thing, hence you would have limit in this case.

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I still do not understand how to approach this question even though I do understand what you just said – Jaden Q Oct 28 '12 at 8:44
@LieX - I don't understand your statement. If the numerator is bounded and the denominator is not, the sequence shouldn't be "bouncing back and forth between finite points" - either it will converge (if the denominator tends to some limit - let it even be $+\infty$ or $-\infty$), or the sequence will not be bounded at all (say, if the denominator is $(-2)^{n}$. – Johnny Westerling Oct 28 '12 at 9:01
Though actually it should converge even in the second case too... What are the other options? – Johnny Westerling Oct 28 '12 at 9:06

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