Tutor.com Infinite Series, Taylor Series Session

Dec. 26, 2012

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Session Transcript - Math - Calculus, 12/23/2012 1:51PM - Tutor.comSession Date: 12/23/2012 1:51PM
Length: 28.2 minute(s)
Subject: Math - Calculus


System Message
[00:00:00] *** Please note: All sessions are recorded for quality control. ***

Richard (Customer)
[00:00:00] I am in the Alternating series portion of infinite series. It is asking me to determine whether the series is convergent or divergent. I will draw the problem, but I take the limit and get infinity/infinity. Does this imply L'Hopital at this point, or does this just mean that because it doesn't equal zero it diverges?

Gabriel I (Tutor)
[00:00:17] Hello, welcome to tutor.com
[00:00:20] lets get started

Richard (Customer)
[00:00:24] ok

Gabriel I (Tutor)
[00:00:50] so i see you are trying to use the fact that the limit as k goes to infinity of each of the terms is not 0

Richard (Customer)
[00:01:16] Right. Is this the correct approach to this problem, or am I confused

Gabriel I (Tutor)
[00:01:24] You can use l'hopitals here on the series of terms

Richard (Customer)
[00:01:45] OK. Let me try it.

Gabriel I (Tutor)
[00:01:47] rather on the absolute value of the terms

Richard (Customer)
[00:03:14] so, i need to use all terms in L'Hopital, or just 3^k/k derivative?

Gabriel I (Tutor)
[00:04:27] Is that what you were asking?

Richard (Customer)
[00:06:19] In a way yes. I wasn't sure what I was asking, but if this is correct than you answered it. Wow. So, derivative of 3^k/k because the (-1)^k+1 is just 1 in absolute value

Gabriel I (Tutor)
[00:06:59] so in order for an alternating series to converge, the absolute values of the terms have to be going to zero right
[00:07:15] and all (-1)^k+1 does is alternate the sign
[00:07:51] so we can disregard it
[00:08:19] So now we get that the limit is infinity/infinity right

Richard (Customer)
[00:09:52] right. So we just proved that in fact it diverges. Correct? I wasn't tracking the absolute value part, but it makes sense. In the working of the problem it denotes oscillation only. The key is it oscillates back and forth closer and closer to zero, not infinity or some other number.

Gabriel I (Tutor)
[00:10:24] no we have not proven it diverges yet
[00:10:39] because infinity over infinity can tend towards 0
[00:11:02] but when we have infinity over infinity or 0 over 0
[00:11:07] we can use l'hopitals
[00:11:35] to get the real limit

Richard (Customer)
[00:11:41] right. we use L'Hopital now. Everythin is so connected to ten sections ago. being able to recall is key.

Gabriel I (Tutor)
[00:11:49] indeed
[00:12:10] do you want to try doing l'hopitals because that top is a little bit tricky

Richard (Customer)
[00:14:33] I am looking at it not sure how to start. Would it be k x 3/1? May need some help here

Gabriel I (Tutor)
[00:15:21] Okay so I'm going to derive a similar one then tell you the general rule
[00:16:16] Does that make sense so far?

Richard (Customer)
[00:16:28] yes

Gabriel I (Tutor)
[00:16:37] now im going to take the derivative of both sides
[00:17:35] is that all clear?

Richard (Customer)
[00:18:32] i see the substitution and looking to get the intuition

Gabriel I (Tutor)
[00:18:58] Can you rephrase your request

Richard (Customer)
[00:19:51] I am trying to process it.

Gabriel I (Tutor)
[00:20:20] Ok in the meanwhile i will give you the general rule for when you just have to deal with the problem
[00:20:29] if you have n a positive real number
[00:20:29] then
[00:21:34] err thats n^xln(n) on the right there
[00:21:56] so lets get back to our problem
[00:22:37] can you fill this in now

Richard (Customer)
[00:23:16] that look about right?

Gabriel I (Tutor)
[00:23:31] great!
[00:23:36] wait
[00:23:41] in the ln there
[00:23:45] you should have ln(3)
[00:24:46] there you go
[00:25:02] now what is the limit as k -> infinity ofg this
[00:25:05] of*

Richard (Customer)
[00:25:58] so this tells me it is still tending to infinity 3^infinity times ln3 is infinity. Correct?

Gabriel I (Tutor)
[00:26:07] yup
[00:26:15] so does this series converge?

Richard (Customer)
[00:26:29] Nope. It is divergent.

Gabriel I (Tutor)
[00:26:39] Great
[00:27:12] Is there anything else I can help you with today

Richard (Customer)
[00:27:36] Thank you for the help. I just started using this today and you all have been a really good christmas present. Have a great holiday!

Gabriel I (Tutor)
[00:27:53] You too, please fill out the survey at the end if you have a second

Richard (Customer)
[00:28:02] will do. take care