Series Convergence

careless25 · 2012-06-02 17:04:52

I am having trouble showing that the below series is convergent and finding its sum:

from n = 1 to inf.

I tried rewriting the series multiple ways but am not able to something i can use. Here is one I think I may be able to use:

Bob · 2012-06-02 18:52:32

hi careless25

That re-arrangement works. It makes both into geometric progressions.

First has 'r' = 1/4 with first term 'a' = 1

Second has r = 1/2 and a = 2

GPs are convergent if |r| < 1

and the sum to infinity is a/(1-r)

Bob

bobbym · 2012-06-02 21:20:41

Hi;

The sum on the right you should recognize as 4.

The other sum,

you say,

Solving we get:

careless25 · 2012-06-03 03:59:03

Hi,

Thanks for your help bob and bobby!

bobbym: I am confused about why you multplied the sum by 1/4 and the steps onwards.

anonimnystefy · 2012-06-03 04:08:35

Hi careless25

When you multiply the sum by 1/4 you get something that looks like your sum,but is smaller by 1.So:

bobbym · 2012-06-03 04:08:51

Hi;

The trick is useful in other types of problems especially generating functions and recurrences. I will not mention the first place people are exposed to it.

You are trying to get part of the above series in terms of Sn. The boxed portion is 1 / 4 of Sn so we can replace it with 1 / 4 Sn. That leaves the equation to solve which gives us the sum.

It is amazing that we can change a sum into a linear equation!

careless25 · 2012-06-03 06:13:26

Oh!
I know of the trick just that the way you wrote it out the first time made it seems like each proceeding step is equal to the previous one.

careless25 · 2012-06-03 08:12:52

I am stuck on another series, I just have to find if its convergent or divergent.

from n = 1 to inf.

I tried doing the Alternating Series Test(Leibniz Test) but I get upto:
lim as n--> inf

and dont know where to go from here.

EDIT: change of sign

Last edited by careless25 (2012-06-03 09:12:29)

anonimnystefy · 2012-06-03 08:25:30

Hi careless25

Check this out:http://www.wolframalpha.com/input/?i=li … %29-n-1%29

It uses the idea of proving divergence using the ratio test.Click the "Show steps" button if you don't know how to get the limit and post here if you have questions.

careless25 · 2012-06-03 08:36:40

I am not sure if thats supposed to help as wolfram says ratio test is inconclusive:

http://www.wolframalpha.com/input/?i=su … D+1+to+inf

anonimnystefy · 2012-06-03 08:43:28

Oh,I copied the problem incorrectly. Yes,the ratio turns out be 1/2 + i/2 or something like that.

careless25 · 2012-06-03 08:46:51

I know what the ratio turns out to be, but I am wondering about how to show that the series diverges or converges. And how to calculate the ratio to show that.

anonimnystefy · 2012-06-03 08:49:35

You cannot use the ratio test here. Did yo try any other convergence tests?

careless25 · 2012-06-03 08:59:03

I got stuck using Alternating Series Test(aka Leibniz Test), cant do Ratio or Root Test and currently I am trying to see what happens to the partial sums of this series.

careless25 · 2012-06-03 09:05:23

Ok I rearranged the expression to this:

when I take the limit:

Last edited by careless25 (2012-06-03 09:10:37)

anonimnystefy · 2012-06-03 09:10:12

Then your signs aren't right in post #8.

careless25 · 2012-06-03 09:14:05

Thanks for pointing that out, but that still doesnt solve the problem. Fixed post #8
and my post above is still correct.

bobbym · 2012-06-03 09:24:16

Hi careless25;

For convergence a series has to first get smaller in absolute value of its terms. Does this one?

careless25 · 2012-06-03 09:43:58

, I have been at this for atleast a couple of hours.
Hmm so all i have to do is show that absolute value of a(n+1) > a(n) ????

bobbym · 2012-06-03 09:46:02

The tests you are using do not apply.

careless25 · 2012-06-03 09:56:45

Yes I know but I am not sure which test would apply now...is this a geometric series? No. Can I use divergence test(n-th term test) on this? I dont think so because this is similar to the case of harmonic series combined with an alternating series(which converges even though the harmonic series diverges by itself).
With the comparision test or limit comparision test...I am not sure what to compare to, -1^n sqrt(n)?
Integral test wont work, this is not a p-series, AST, ASET, root and ratio all dont work.

Therefore by deduction LCT or Comparision Test should do it!

bobbym · 2012-06-03 10:02:27

The series diverges. Try to prove that.

careless25 · 2012-06-03 11:24:23

I have given up on this for today, will look at it with a fresh mind tomo.

bobbym · 2012-06-03 16:13:17

Hi careless25;

Good idea! I am going to go eat.

anonimnystefy · 2012-06-04 06:48:22

Hi careless25

Last edited by anonimnystefy (2012-06-04 17:11:00)

Math Is Fun Forum

#1 2012-06-02 17:04:52

Series Convergence

#2 2012-06-02 18:52:32

Re: Series Convergence

#3 2012-06-02 21:20:41

Re: Series Convergence

#4 2012-06-03 03:59:03

Re: Series Convergence

#5 2012-06-03 04:08:35

Re: Series Convergence

#6 2012-06-03 04:08:51

Re: Series Convergence

#7 2012-06-03 06:13:26

Re: Series Convergence

#8 2012-06-03 08:12:52

Re: Series Convergence

#9 2012-06-03 08:25:30

Re: Series Convergence

#10 2012-06-03 08:36:40

Re: Series Convergence

#11 2012-06-03 08:43:28

Re: Series Convergence

#12 2012-06-03 08:46:51

Re: Series Convergence

#13 2012-06-03 08:49:35

Re: Series Convergence

#14 2012-06-03 08:59:03

Re: Series Convergence

#15 2012-06-03 09:05:23

Re: Series Convergence

#16 2012-06-03 09:10:12

Re: Series Convergence

#17 2012-06-03 09:14:05

Re: Series Convergence

#18 2012-06-03 09:24:16

Re: Series Convergence

#19 2012-06-03 09:43:58

Re: Series Convergence

#20 2012-06-03 09:46:02

Re: Series Convergence

#21 2012-06-03 09:56:45

Re: Series Convergence

#22 2012-06-03 10:02:27

Re: Series Convergence

#23 2012-06-03 11:24:23

Re: Series Convergence

#24 2012-06-03 16:13:17

Re: Series Convergence

#25 2012-06-04 06:48:22

Re: Series Convergence

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