Question

Help please:
Does this series CONVERGE/DIVERGE?

sum(1->infinity) ((-1)^n)*(n)/(sqrt((n^3) + 2))

This is in the topic of Alternating Series Test.
I used wolfram and the Ratio and the Root test are inconclusive. So, no choice but Alternating Series.

And so...I tested the first assumption that:
b_(n+1) <= b_(n)
which is False, because when I plugged in 1, it is:

6<=5
And so it fails the first test, thus DIVERGENT.
My question is, is my conclusion correct? I'm not sure.

Answer 1

Note that while it is correct that you need:\[b_{n+1}\le b_n\]and it is also correct that when you plug in n = 1, the statement is false, what you are actually looking for is that the sequence is eventually decreasing. It doesnt matter if it increasing from n = 1 to n = whatever, as long as its decreasing for all n after that. Your problem has that property. You can use derivatives to see that it is eventually decreasing.

Answer 2

Okay...
I also tried that approach but I could not find anything:
f'(x) < 0
As in there is no x that would make f'(x) < 0
so...does that mean it is not decreasing?

Answer 3

Or you can note that:\[\frac{n}{\sqrt{n^3+2}}\le \frac{n}{n^{\frac{3}{2}}}=\frac{1}{n^{\frac{1}{2}}}\]and the latter is a eventually decreasing.

Answer 4

ok...

1/(n^(1/2)) is a p-series with p = 1/2 and <1, and so it Diverges.
But if you put it in a comparison like that, it does not say anything good, because 1/(n^(1/2)) is greater than the other, and not less than the other.

Answer 5

Im not trying to do a comparison test, I'm only try to show that \[\frac{n}{\sqrt{n^3+2}}\]is a decreasingg sequence as some point.

Answer 6

Oh I see....
You're not saying to use the comparison test.
You're just saying that the function will eventually decrease because..

Answer 7

Ok...but is there any other way to prove that it is decreasing, without giving that thing?

Answer 8

The derivative should have signaled the decreasing nature. Let me compute it real fast to see if thats the case.

Answer 9

It won't give you any value of x that would show f'(x) < 0

Answer 10

\[\frac{d}{dx}\left(\frac{x}{(x^3+2)^{\frac{1}{2}}}\right)=\frac{(x^3+2)^{\frac{1}{2}}-x(\frac{1}{2}(x^3+2)^{-\frac{1}{2}}(3x^2))}{x^3+2}\]Multiply the top and bottom of the fraction by:\[(x^3+2)^{\frac{1}{2}}\]to get:\[\frac{(x^3+2)^{\frac{1}{2}}-x(\frac{1}{2}(x^3+2)^{-\frac{1}{2}}(3x^2))}{x^3+2}=\frac{x^3+2-\frac{3}{2}x^3}{(x^3+2)^{\frac{3}{2}}}\]

Answer 11

As x gets large, that will become negative at some point.

Answer 12

if you have the f'(x)
you might equate it to < 0
f'(x) < 0

which will give you

(x^3 + 2) - (3/2)x^2 < 0
(x^3 + 2) < (3/2)x^2
2x^3 + 4 < (3x^2)

Answer 13

Okay. Thank you.

Answer 14

I found the answer. I asked the teacher about this and he told me that I could just say that f'(x) is decreasing as x->infinity.

Thanks again. :)