Infinite series question

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  • #1
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I have to show that ln(1 - 1/k^2)= - ln(2)

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

How should i proceed now?
 

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  • #2
dextercioby
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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 have to show that ln(1 - 1/k^2)= - ln(2)

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 Difference of Two Squares: a2 - b2 = (a - b) (a + b) right? Maybe that should help.
 
  • #5
dextercioby
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Get rid of the natural logarithms. You'll get an infinite product in the LHS and a simple fraction in the RHS. From then on it's trivial.
 
  • #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 the Difference of Two Squares: a2 - b2 = (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 n first, and then take the limit as n grows without bound.
 
  • #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 to n first, and then take the limit as n grows 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 n first, then take the limit as n tends 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.
 
  • #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 to n first, then take the limit as n tends 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 n first. There are 2 more terms that are not canceled out. And then, take the limit as n tends to infinity.
 
  • #12
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First, as I mentioned above, you should try summing to n first. There are 2 more terms that are not canceled out. And then, take the limit as n tends to infinity.

Finally I get it! Thanks alot dude!
 

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