This was a question i very much enjoyed answering because many people have trouble with - and actually absolutely hate! - power series and convergence and whatnot. but it is exactly the opposite! i LOVED the subject! (don't even get me started on Taylor Series!) and so i feel that this is an area in which i can do a lot of good and help clarify some subjects that some people have trouble with and dislike for no good reason at all!
and i'll also say that one of things that i think makes this subject hard for some people is that it requires the ability to really understand the mathematics and be able to draw conclusions and meaning from the equations. this is something A LOT of people lack, and it is not even something that is emphasized in schools! and this is a terrible circumstance, which i battle frequently, and i hope that i am doing my part to give people this familiarity with mathematics which is really useful and, above all, FUN!
it is also good to see that the asker was very pleased by my answer, and i really hope i was able to make a difference in the way he/she thinks about these subjects.
Q:
Calculus: Power Series?
I thought that determining weather series converged or diverged was easy, but I seem to be having some problems figuring out how to do that with power series, so can you help me with this problem. Thanks :)
Find the interval of convergence for the given power series:
∑ (n=1 to ∞) [((x-9)^n) / (n(-7)^n)]
Additional Details [the asker posted this after someone else had posted a response with just the answer and nothing else. really pathetic in my opinion...]
I already knew that the answer was (2,16] so you haven't helped me at all. I want to see the steps taken to get to the answer because that is the part that I need help with.
A:
the key you need to solving this problem is to use the ratio test. most of the time, the ratio test will get you the answer you need to convergence problems.
if you recall, the ratio test is the limit (as n approaches ∞) of the absolute value of the ratio of the n+1th term to the nth term of the sequence. or in a different language:
lim(n-->∞) | (a_n+1) / (a_n) |
in this case, the nth term of your sequence is a_n = (x-9)ⁿ / n*(-7)ⁿ.
it follows that the next term (the n+1th term) is a_n+1 = (x-9)ⁿ⁺¹ / (n+1)*(-7)ⁿ⁺¹.
it also happens that our sequence is a function of another variable (x), but this is of no consequence when we consider a limit whose parameter is n (and n and x are in no way related).
so now we need to take the ratio of these as n goes to infinity. we write:
lim(n-->∞) | [ (x-9)ⁿ⁺¹ / (n+1)*(-7)ⁿ⁺¹ ] / [ (x-9)ⁿ / n*(-7)ⁿ ] |
this looks ugly, so we simplify the fractions:
lim(n-->∞) | [ n*(x-9)ⁿ⁺¹ (-7)ⁿ ] / [ (n+1) (x-9)ⁿ (-7)ⁿ⁺¹ ] |
as you can see, some nice things cancel here (since Aⁿ⁺¹ = Aⁿ A¹)
[at this point i highly suggest you write this out as you read it because it become much easier to read and see for yourself what i mean]
now that we've cancelled, our limit looks as follows:
lim(n-->∞) | n*(x-9) / (n+1)*(-7) |
now, if we wish (and we do), we can take things which do not depend on n out of the limit.
notice that x-9 does not depend on n (it only depends on x) and -7 doesn't really depend on anything (especially n!) so we can take these both out of the limit. but the absolute value signs stay on! (don't forget those!)
so, our limit now looks like this:
|(x-9)/(-7)| lim(n-->∞) | n / (n+1) |
the next step is to evaluate this rather simple limit. note also that you can just take those absolute value signs right off of the n/(n+1). this is because everything inside them is guaranteed to be positive, so the signs are superfluous. (but don't take them off of the (x-9)/(-7)! those aren't guaranteed to be positive!)
so, we know that n/(n+1) approaches 1 as n approaches infinity, and so the value of the limit is 1.
our expression now looks simply like:
|(x-9)/(-7)|
now, i've been neglecting it, but the point of the ratio test is to see of the value of the ratio is less than 1 because this is where the series converges (and don't forget that it is inconclusive when it equals 1!)
so we really have to solve this problem:
|(x-9)/(-7)| < 1
also note that that -7 on the bottom can be written as just a 7 and taken out of the absolute value signs. (because |-7| = 7)
so now let's solve
|x-9| / 7 < 1
|x-9| < 7
as you know, you can write this 2 ways and get rid of the absolute value signs as:
x-9 < 7, -(x-9) < 7
solving these (finally), we get:
x < 16, x > 2
(don't forget that when you multiply by a negative in an inequality, you flip the sign)
so we're done right? almost. remember that the ratio test inconclusive when it equals 1. so we have to manually check those possibilities. we do this by checking convergence of the original series at a particular value of x. recall the original series:
∑ [ (x-9)ⁿ / n*(-7)ⁿ ]
the 2 we have to check now are x = 16 and x = 2. notice that at x = 2, the (-7)ⁿs cancel on the top and bottom and you are left with 1/n. as you should know, ∑ 1/n diverges. so, x = 2 does not yield a series which converges.
at x = 16, you get a different series. i'll actually leave it to you to check this one for convergence (hint: use the ratio test again). but if you have problems with it, just let me know, and i'll walk you through it again.
i hope this has helped! and i'm always happy to answer more questions if you have any :]
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