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-   -   Infinite series- converges or not (http://www.physicsforums.com/showthread.php?t=210321)

asif zaidi Jan22-08 11:58 AM

Infinite series- converges or not
 
I have 2 questions I am having problems with.
The goal is to determine if the series converges or not.


Q1: Sum from(1 to inf) of (exp^i)/( (exp^2i) + 9)


I tried to do the integral test but I cannot seem to integrate. Any guides would be appreciated.

Also if I wanted to compare it what would I compare it to.




Q2: Sum from (2 to inf) of 1/( (i^2) - 1)

I did the integral test and came out to the following

integral of 1/i^2 -1 = 1/2 (ln(i-1) - ln(i+1))

So in looking at this as i approaches infinity, can I say that integral approaches 0 and therefore this series converges.

All help is appreciated.

Thanks

Asif

Dick Jan22-08 12:54 PM

There's an easy comparison for both of them. Q1) compare to exp(i)/exp(2i), Q2) compare to 2/(i^2).

Tom Mattson Jan22-08 12:55 PM

Quote:

Quote by asif zaidi (Post 1580435)
I have 2 questions I am having problems with.
The goal is to determine if the series converges or not.


Q1: Sum from(1 to inf) of (exp^i)/( (exp^2i) + 9)


I tried to do the integral test but I cannot seem to integrate. Any guides would be appreciated.

Also if I wanted to compare it what would I compare it to.

The integral test is indeed a good way to go. Can you show us how you tried to integrate it? Once you do that, I'll point you in the right direction.


Quote:

Q2: Sum from (2 to inf) of 1/( (i^2) - 1)

I did the integral test and came out to the following

integral of 1/i^2 -1 = 1/2 (ln(i-1) - ln(i+1))

So in looking at this as i approaches infinity, can I say that integral approaches 0 and therefore this series converges.
The series does converge, but you can do better than the integral test. If you recognize that the series is telescoping, you can actually find its sum.

asif zaidi Jan22-08 01:11 PM

Dick: thanks for response. I will look at your way. Always good to learn new ways

Tom: This is how I did the integral test

Let u = exp^(2x) + 9
du/dx = 2exp^(2x)
This is where I got stuck. I thought once I do du/dx, I should get e^x in this equation but since I didn't I couldn't get anywhere.

So I tried by integration by parts and went into an infinite loop

a- let u = 1/( exp(2x) + 9)
b- du/dx = -2/(exp(2x +9)
c- let dv = exp^x dx
d- v = e^x

To integrate
uv - integral (v wrt dx)
integral (e^x/e^2x +9) = e^x/(e^2x + 9) +2 integral (e^x/e^2x +9)

This is where I got stuck.

Thanks

Asif

Tom Mattson Jan22-08 01:21 PM

Quote:

Quote by asif zaidi (Post 1580510)
Dick: thanks for response. I will look at your way. Always good to learn new ways

Tom: This is how I did the integral test

Let u = exp^(2x) + 9

You want to let u=exp(x) instead. You will end up with a rational function in u whose antiderivative you should recognize.


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