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October 2nd, 2016, 09:29 AM   #1
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Series, did I solve this correctly?

Also I wasn't sure if it'd be correct to evaluate the limit from i = 3 since that's where the sum starts or if I'd ignore that and do it as if it was i = 1 anyways. I'm not really sure if this is even close to right or not to be honest.

you can click on the picture to make it bigger, I'm not sure why it came out so small on here.
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October 2nd, 2016, 09:57 AM   #2
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No.

You can factorise your first denominator ${4i^2-1}$ further to get three terms in your sum.

Once you have done that, you can write out a few terms. You should see that you get a telescoping sum.
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October 2nd, 2016, 10:47 AM   #3
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Like this?

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October 2nd, 2016, 11:09 AM   #4
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I just realized I may have misread the problem. the problem was
"find the sum as a function of n" I'm guessing like how the sum of i^2 integers is (n(n+1))/2
except with this. Would it work if I split the fractions and found the sum as a function of n for the individual fractions?

either way it was good to practice evaluating the sum, I need to practice that more.
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October 2nd, 2016, 12:57 PM   #5
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Quote:
Originally Posted by GumDrop View Post
I just realized I may have misread the problem. the problem was
"find the sum as a function of n" I'm guessing like how the sum of i^2 integers is (n(n+1))/2.
Your penultimate line is a function of $n$, isn't it?
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Last edited by v8archie; October 2nd, 2016 at 12:59 PM.
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October 2nd, 2016, 01:12 PM   #6
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Quote:
Originally Posted by v8archie View Post
Your penultimate line is a function of $n$, isn't it?
lol, I feel silly. Thanks v8archie. Just to make sure I did end up getting the correct answer right? I'm going to do a few more problems like this and I want to make sure I'm not missing anything
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October 2nd, 2016, 01:19 PM   #7
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I haven't checked your algebra, but you appear to have followed the right steps.
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Last edited by skipjack; November 8th, 2016 at 05:40 PM.
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October 3rd, 2016, 07:16 AM   #8
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What about this one? (Sorry, but there's no answer for these and I want to make sure I understand this stuff)



Also, I put the series approaches 0 as n approaches infinity, but I think it'd be that the series approaches negative infinity as n goes to infinity.
(I forgot bout the -ln(n) as I was writing the answer)

Thanks for the help as always btw.
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October 3rd, 2016, 07:34 AM   #9
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I would deal with the root by noticing that $\frac{(4n)!}{(2n)!}$ is a product of $2n$ terms each of which is between $2n$ and $4n$ and that we can therefore say that

$$(2n)^{2n} < \frac{(4n)!}{(2n)!} < (4n)^{2n}$$

You can use one of these inequalities to run a comparison test.
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October 3rd, 2016, 08:07 AM   #10
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You mean like this?



I've never really done a comparison test before, I had to google it so I'm not really sure if I did what you meant, but I got a different answer this time. before it was -infinity now it's +infinity

I just realized the limits in the picture for post #3 weren't right. I put that was n -> infinity 1/(2n + 1) = 1/2 ... oops

Last edited by GumDrop; October 3rd, 2016 at 08:41 AM.
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