Question

Does this series conditionally converge, absolutely converge, or diverges

(-1)^(n-1) (3^n)/(n^4)

Answer 1

\[\huge \sum_{n=1}^{\infty}(-1)^{n-1}\frac{3^n}{n^4}\]
Remember that for a series to be convergent, it is necessary for the underlying sequence to converge... to zero!

Answer 2

It does! By alternating series test it converges. Because it converges it would be conditionally convergent? So I wouldn't have to check if it absolutely converges. I tried using ratio test but I got the limit to equal 1 which is inconclusive

Answer 3

I meant to ask, would I have to check with absolute test?

Answer 4

I find that hard to believe :)
\[\huge 3^n\] is exponential, which increases, well, exponentially
\[\huge n^4\] is just a polynomial, which increases, but not nearly as quickly as an exponential.

I don't think
\[\huge \frac{3^n}{n^4}\] goes to 0 :)

Answer 5

If you're not convinced, try evaluating this limit...
\[\huge \lim_{x \rightarrow \infty}\frac{3^x}{x^4}\]
Use L'hopital's rule.

Answer 6

Well I was thinking, if the bottom was reaching infinity faster than itd be zero. Like 1/infinity. But maybe that logic is very wrong so I'll try lhosp rule

Answer 7

You're right!
If the bottom reaches infinity faster, it goes to zero.

But it's the TOP (3^n)
that goes to infinity faster :)

Answer 8

Ok yeah I tried lhosp. But the 3^n would continue be ln 3 *3^n. and I'd get #/0 after using lhos 4 times

Answer 9

Yes... so it goes to infinity, right? :)
Not zero, at least, since it's some number divided by 0.

Answer 10

Yes!

Answer 11

So if it were to converge using an absolute convergence test such as ratio test, then I wouldn't have to see if the original series diverges? I can just claim it as absolutely convergent?

Answer 12

Thank you!

Answer 13

Wait, what? For the record, this series is divergent.

Answer 14

Yes! I'm just asking a general question

Answer 15

Oh, okay :)
If it passes the ratio test (or the root test) then it's absolutely convergent.

By pass, I mean the limit you compute is less than 1. :)

Answer 16

But to find if its conditionally convergent, ratio/root test must diverge and the original series must converge?

Answer 17

@kimmy0394
Conditionally convergent just means it's convergent, but not absolutely convergent :)
Case in point, the classic Harmonic Series that alternates...
\[\huge \sum_{n=1}^{\infty}(-1)^{n+1}\frac{1}{n}\]

Answer 18

oh ok thanks!

Answer 19

Here's a tip...
If the sequence converges to zero, it's AT LEAST conditionally convergent, as a series.
It MIGHT be absolutely convergent, you might want to use something like ratio or root test for that.

Answer 20

ok i'll keep that in mind :)