Does the sequence converge, or diverge?
(-1)^(n-1)/sqrt(n)

Question

Does the sequence converge, or diverge?
 (-1)^(n-1)/sqrt(n)

anonymous · Answer

From 1 to infinity.

anonymous · Answer

oop it is a SEQUENCE right, not a SERIES

anonymous · Answer

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

anonymous · Answer

as $n\to \infty$ it should be pretty clear that $a_n\to 0$

anonymous · Answer

How though? That's what I don't understand..The steps to get that answer. @satellite73

anonymous · Answer

I would suggest showing that the sequence is convergent by showing it is both monotonic and bounded.

Since you have an alternating series, the monotonicity part might be tricky. As a hint, try working with the positive and negative terms separately.

anonymous · Answer

@SithsAndGiggles any idea to start it?

anonymous · Answer

$$\{a_n\}=\left\{\frac{(-1)^{n-1}}{\sqrt{n}}\right\}_{n=1}^\infty=\left\{1,-\frac{1}{\sqrt2},\frac{1}{\sqrt3},-\frac{1}{2},\cdots\right\}$$
Consider $\{b_n\}=\left\{\dfrac{(-1)^{2k-1}}{\sqrt{2k}}\right\}_{k=1}^\infty$ and $\{c_n\}=\left\{\dfrac{(-1)^{2k}}{\sqrt{2k+1}}\right\}_{k=0}^\infty$. Notice that $b_n$ is the subsequence of $a_n$ consisting of the even terms ($a_2,a_4,\cdots$) and $c_n$ is the subsequence of the odd terms ($a_1,a_3,\cdots$).

Showing that $b_n$ and $c_n$ converge should be enough to show that $a_n$ converges.

Alternatively, you can show the following conditions hold:
(1) the magnitude of each term is decreasing, meaning $|a_n|\ge|a_{n+1}|$, and
(2) the terms converge to 0.

I think the second method would be a more valid approach. The first is a bit more hand-wavy.

anonymous · Answer

So I should just prove that An converges?

anonymous · Answer

$a_n$ is what you're given, so yes, that's what you want to show convergence for. Show that the conditions are met and you're done.