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