Interval of Convergence Series x^2n/(-3)^n

Kuli02

New member
Joined
Jul 20, 2021
Messages
2
I stuck with this problem.
Find the interval of convergence of
(x^2n)/((-3)^n)

I already found out that the boundaries of this Interval are [math]-{\sqrt{3}}[/math] and [math]{\sqrt{3}}[/math].
When I try to check the boundaries Insertertet, I get the soultion, that both have to be excluded. But one Series calculator in the Internet said, [math]-{\sqrt{3}}[/math] has to be included. That confuses me, because when I insert this boundary, i get the same series as when I insert the other boundaries.
I am very thankful for any advises!
 

Zermelo

Junior Member
Joined
Jan 7, 2021
Messages
81
Hi there, if you substitute x for [imath]- \sqrt{3}[/imath] in the series, what real number series do you get as a result? What do we know about series of real numbers, precisely, what is the necessary condition for convergence of a real number series? Does the given series satisfy that condition? If no, then the series definitely doesn't converge! If yes, you need to do some convergence tests.
I myself think that the online series calculator is wrong, but feel free to check it yourself!
 

Kuli02

New member
Joined
Jul 20, 2021
Messages
2
Thanks for your reply! When I put both results in i get the series (-1)^n which does neither converge or diverge. So I would exclude the boundaries. When I try values inside the boundaries, it works, because then we have an alternating geometric series. Good to hear, that you also think there is a mistake in the online tool
 

lex

Full Member
Joined
Mar 3, 2021
Messages
842
Yes, clearly you get the same series with terms [imath](-1)^n[/imath] (n is an integer).
 
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