Geometric Series

  • Thread starter Rossinole
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  • #1
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Homework Statement



Does the series from n=1 to infinity of (2)/(n^2-1) converge or diverge? If it converges, find the sum.

Homework Equations





The Attempt at a Solution



I can see right away that the series converges by a limit comparison test by looking at the series. However, to find the sum I have re-write that as a geometric series. There is nothing, at least to me, that gives away how to re-write that as a geometric series. That's where I'm stuck.

Thanks for any help.
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
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Why don'y you write [itex]\frac{2}{n^2-1}[/itex] in partial fractions and see if it is a telescoping series?


Find

[tex]\sum_{n=1} ^{N} \frac{2}{n^2-1}[/tex]

and then check what happens as [itex]N \rightarrow \infty[/itex]
 
  • #3
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I see it now. Thank you.
 
  • #4
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what's the summation of the series?

what's the summation of the series?
1.A^1 + 2.A^2 + 3.A^3 + .............+n.A^n.

please reply
 
  • #5
rock.freak667
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what's the summation of the series?
1.A^1 + 2.A^2 + 3.A^3 + .............+n.A^n.

please reply

What have you tried so far on it?
 
  • #6
Defennder
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what's the summation of the series?
1.A^1 + 2.A^2 + 3.A^3 + .............+n.A^n.

please reply
Is this related to the original post or something separate?
 
  • #7
statdad
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Are you sure you don't mean

[tex]
\sum_{n=2}^\infty \frac{2}{n^2-1}
[/tex]

i.e.- the sum starting at [tex] n = 2 [/tex]? If you try to start at [tex] n = 1 [/tex] the very first term is undefined (can't divide by zero) so the series would not converge. As I mentioned - this makes the difference between the series converging and not converging, and will influence your value for the sum.
 
  • #8
Gib Z
Homework Helper
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Yes, he probably meant the sum to start at n=2.

Jainal36- you should really start new threads rather than hijacking others, but ill help anyway - thats just A times the derivative of a series you know how to sum.
 

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