Root Test for Series

redrum41987 Says:

Feb 21, 2012 - gotta love those ay-su-binz

khilozozo02 Says:

Mar 4, 2012 - i dont understand why there are any dislikes. who are these people??

BER2ERKER Says:

Mar 8, 2012 - @kenikozo Like an "n" in exponent right? That is a geometric series. I don't know if "non-geometric" series can have an exponent of n, maybe it's possible; but they'll likely always be geometric, ^n. I'm sure Patrick would correct me if I'm wrong.

BER2ERKER Says:

Mar 8, 2012 - I thought series were going to be hard, but this whole concept so far, including doing differential eqtns with power series, and the Taylor polynomial section, has been easier than integration techniques IMO. First time in a while that I didn't actually need Patrick's videos to get it down correctly. Still watched them though, they always have something inventive to teach you.

BER2ERKER Says:

Mar 8, 2012 - @kenikozo Oh sorry, you said limit, not series. I don't know then. Because if you have a variable in a basic limit going to infinity, that variable is what would be replaced with infinitely increasing numbers. The only thing I could think of is a constant to the x power, like 2^x, where x approaches inf. Though, I probably am forgetting something. Anything to the "infinite power" would be increasing or decreasing without bound though I'm pretty sure.

mastamind9682 Says:

Mar 25, 2012 - You are a gentleman and scholar.

heresatissue Says:

Mar 29, 2012 - What happened to the exponent n attached to the 1? It dissappeared after the third to last step? How does (1^n)/[(1+1/n)^n]=(1)/[(1+1/n)^n]?

heresatissue Says:

Mar 29, 2012 - You said 1^n=1, nvm.

jojaba2u Says:

Apr 10, 2012 - wtf is up with the cap on your sharpie lol

usman2hype92 Says:

Apr 10, 2012 - AN EASY WAY TO REMEMBER MAJOR RULES/TESTS: ABBREVIATED DARRIC.D: Divergence TestA: Alternating Series TestR: Ratio TestR: Root TestI: Integral TestC: Comparison TestLIKE so people can remember them for their tests/exams :)

AndyTemps Says:

Apr 13, 2012 - In the second example, why did he (seemingly randomly) multiply the numerator and denominator by 1/n?

BeeryGamer Says:

Apr 15, 2012 - What would happen if there were a n^n on the outside of a series, how would a root test work on that?

Vakalao79 Says:

Apr 17, 2012 - Thats how you do the ratio test

AnthonyPickett Says:

Apr 18, 2012 - wow. im at UT too.. and the whole time I wondered where patrick was at and it was right here.. same world!! Just show Austin rules!

AnthonyPickett Says:

Apr 18, 2012 - small world*

XxMANKITTYXx Says:

Apr 19, 2012 - if i wasn't a man, i would ask you to father my children

RealLilSmarty Says:

Apr 22, 2012 - how about limit comparison test ??

ICarnag3I Says:

Apr 29, 2012 - Hate series. Don't think it has many applications to engineering.

408Nano69 Says:

Apr 30, 2012 - due you mean why he raised it to the (1/n)? and if thats the question its because the equation has you find the limit of [an] to the nth root which is the same as raising the problem to (1/n). an example is the the square root of 4 which can be written as 4^(1/2). hope this helps :)

alexatlanta1983 Says:

May 1, 2012 - good way to remeber the tests (R.eally R.eally A.mazing D.octor In C.anada)D: Divergence TestA: Alternating Series TestR: Ratio TestR: Root TestI: Integral TestC: Comparison Test

deepintheslums Says:

May 2, 2012 - What is this supposed to help you with? If you know the methods to the tests you should be able to remember the names of them...

khwezi1986 Says:

May 3, 2012 - I think you made a mistake on your second example your denominator is 0 therefore you have infinity / 0 not infinity / 27 and this implies L'Hopital

JnCrWe Says:

May 8, 2012 - The exponent 1/n goes to 0, so 3^(1/n) goes to 3^0 = 1. The denominator therefore goes to 1*(3^3) = 1(27) = 27. :)

SomethingSoOriginal Says:

May 24, 2012 - Thanks, no-one does analysis on youtube as well as you do!