Geometric Series Homework: Converge or Diverge? Find Sum

In summary, the conversation discusses the convergence of the series from n=1 to infinity of (2)/(n^2-1) and how to find its sum. The series is shown to converge using a limit comparison test and can be rewritten as a geometric series. However, there is difficulty in finding the sum until the suggestion to use partial fractions and a telescoping series method is given. The conversation then briefly touches on a separate series before concluding with a reminder to start new threads.
  • #1
Rossinole
20
0

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.
 
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  • #2
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
I see it now. Thank you.
 
  • #4
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


jainal36 said:
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


jainal36 said:
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
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
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 - that's just A times the derivative of a series you know how to sum.
 

1. What is a geometric series?

A geometric series is a sequence of numbers where each term is found by multiplying the previous term by a constant number called the common ratio. It is written in the form of an = a1 * r^(n-1), where a1 is the first term and r is the common ratio.

2. How do you determine if a geometric series converges or diverges?

A geometric series converges if the absolute value of the common ratio is less than 1, as the terms will get smaller and smaller as n increases. It diverges if the absolute value of the common ratio is greater than or equal to 1, as the terms will get larger and larger as n increases.

3. What is the sum of a convergent geometric series?

The sum of a convergent geometric series can be found using the formula Sn = a1 * (1 - r^n)/(1 - r), where Sn is the sum of the first n terms, a1 is the first term, and r is the common ratio.

4. How do you find the sum of a divergent geometric series?

A divergent geometric series does not have a finite sum. However, if the common ratio is between -1 and 1, the series can be converted into a convergent series by taking the limit as n approaches infinity.

5. Can a geometric series have a negative common ratio?

Yes, a geometric series can have a negative common ratio. This will result in alternating positive and negative terms in the series. The convergence or divergence of the series will still be determined by the absolute value of the common ratio.

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