The Convergence of an alternating series test

Question

Can I confirm that $$\sum \frac{(-5)^{n}}{n^{3}}$$ converges by the alternating series test?

No. Dirichlet conditions are violated. – Landon Carter Mar 27 '15 at 10:48 — Landon Carter, Mar 27 '15 at 10:48

score 1 · Answer 1 · answered Mar 27 '15 at 08:33

1

HINT: $$5^n\gt n^3,\,\,\,\forall n\in\mathbb{N}$$

answered Mar 27 '15 at 08:33

Bumblebee

It tends to zero as n tends to $$\infty$$ – guest Mar 27 '15 at 08:43
1

NO: $\lim_{n\to\infty}\dfrac{5^n}{n^3}\ge 1.$ – Bumblebee Mar 27 '15 at 08:46
How is it possible? Mathematica reaffirms that the limit is infinity.. – guest Mar 27 '15 at 08:48
Of cause, $\lim_{n\to\infty}\dfrac{5^n}{n^3}=\infty\not= 0.$ Therfore you can not apply the AST for this series. – Bumblebee Mar 27 '15 at 08:50
2

You can use L'Hospital's Rule to evaluate this limit. – Bumblebee Mar 27 '15 at 08:57

peter.petrov · Answer 2 · 2015-03-27T09:14:33.903

0

This series doesn't converge because its n-th member
doesn't go to $0$ as $n$ goes to infinity. And it doesn't
go to $0$ because:

$|\frac{(-5)^n}{n^3}| >= 1 $ for all n

So you cannot confirm (by any test) that it converges as that's not true.

edited Mar 27 '15 at 09:14

answered Mar 27 '15 at 09:08

peter.petrov

My main problem with the question on hand is having to determine whether this series converges absolutely and if it does not, what. – guest Mar 27 '15 at 09:17
What I had tried to do was to use the alternating series test and letting b_n = 1/n^(3). Then by applying the limit as n tends to infinity, the term of the series b_n tends to 0. – guest Mar 27 '15 at 09:19
@guest It is even easier than what you try to prove: the given series does not converge at all : not absolutely, not conditionally...nothing. This, and also the other, answer explains why. – Timbuc Mar 27 '15 at 09:29
I see it! Thank you! Easy peasy – guest Mar 27 '15 at 09:45

2 Answers2