- #1

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I took LCM and separated the equation using ln(A/B) = lnA - lnB

How should i proceed now?

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- Thread starter jokerzz
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- #1

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I took LCM and separated the equation using ln(A/B) = lnA - lnB

How should i proceed now?

- #2

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?? Can you state your problem correctly ? Some summing or product symbols are definitely missing.

- #3

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?? Can you state your problem correctly ? Some summing or product symbols are definitely missing.

sigma k=2 to infinite is also there

- #4

VietDao29

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I took LCM and separated the equation using ln(A/B) = lnA - lnB

How should i proceed now?

You describe your process too vaguely. But yes, you seem to be on the right track. :) Can you show us where you get to, so that we can check your work, and guide you from there?

You know the

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- #6

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You describe your process too vaguely. But yes, you seem to be on the right track. :) Can you show us where you get to, so that we can check your work, and guide you from there?

You know theDifference of Two Squares:a^{2}-b^{2}= (a-b) (a+b) right? Maybe that should help.

Ok so now the equation is: ln(k-1)+ln(k+1) - 2ln(k)

so now what do i need to do? Apply the limit k-> infinity? But that wouldn't prove anything

- #7

VietDao29

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Ok so now the equation is: ln(k-1)+ln(k+1) - 2ln(k)

so now what do i need to do? Apply the limit k-> infinity? But that wouldn't prove anything

It's summation, so there should be some terms you can cancel out. :)

You can try summing from k = 2 to

- #8

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It's summation, so there should be some terms you can cancel out. :)

You can try summing from k = 2 tonfirst, and then take the limit asngrows without bound.

Im sorry but my concepts in this topic arnt very clear. Can you please elaborate a little?

- #9

VietDao29

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Ok so now the equation is: ln(k-1)+ln(k+1) - 2ln(k)

so now what do i need to do? Apply the limit k-> infinity? But that wouldn't prove anything

Your series would look something like:

[tex]\sum_{k = 2} ^ {\infty} \left[ \ln(k - 1) + \ln(k + 1) - 2 \ln (k) \right][/tex], right?

Instead of summing to infinity, you should sum from k = 2 to

[tex]\lim_{n \rightarrow \infty}\left\{ \sum_{k = 2} ^ {n} \left[ \ln(k - 1) + \ln(k + 1) - 2 \ln (k) \right] \right\}[/tex]

You can write out some of the first terms to see the pattern.

ln(1) + ln(3) - 2ln(2) + ln(2) + ln(4) - 2ln(3) + ...

What should you do now is to try to cancel out the terms. Hint: Most of them can be canceled out.

- #10

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Your series would look something like:

[tex]\sum_{k = 2} ^ {\infty} \left[ \ln(k - 1) + \ln(k + 1) - 2 \ln (k) \right][/tex], right?

Instead of summing to infinity, you should sum from k = 2 tonfirst, then take the limit asntends to infinity. What I mean is this:

[tex]\lim_{n \rightarrow \infty}\left\{ \sum_{k = 2} ^ {n} \left[ \ln(k - 1) + \ln(k + 1) - 2 \ln (k) \right] \right\}[/tex]

You can write out some of the first terms to see the pattern.

ln(1) + ln(3) - 2ln(2) + ln(2) + ln(4) - 2ln(3) + ...

What should you do now is to try to cancel out the terms. Hint: Most of them can be canceled out.

Yeah all the terms accept -ln2 are cancelling out. But isn't this method a little crude? Isn't there any other way to prove it?

- #11

VietDao29

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Yeah all the terms accept -ln2 are cancelling out. But isn't this method a little crude? Isn't there any other way to prove it?

First, as I mentioned above, you should try summing to

- #12

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First, as I mentioned above, you should try summing tonfirst. There are2 more termsthat arenotcanceled out. And then, take the limit asntends to infinity.

Finally I get it! Thanks alot dude!

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