
Junior Member
Determine if series convergent/divergent. Find its sum
Hello,
I need to:
Determine if series is convergent or divergent. If convergent, Find its sum.
I have a problem: Summation from n=2 to infinity of: 2/(n^2  1)
The limit as n> infinity of 2/(n^2  1) is 0, so the divergence test fails.
I think what i'm left to work with is whether it's a harmonic, telescoping, or geometric series.
If I drop the index of the summation from n=2 to n=1, 2/(n^2  1) becomes 2/[(n + 1)^2  1] I believe... Then if I evaluate it:
a1 = 2/3
a2 = 2/3 + 1/4
a3 = 2/3 + 1/4 + 2/15
a4 = 2/3 + 1/4 + 2/15 + 1/12
Is this correct? I'm not sure how to tell if this is convergent/divergent from this. I don't recognize this as a harmonic, telescoping, or geometric sequence... any ideas?

Elite Member
Re: Determine if series convergent/divergent. Find its sum
Have you looked at the Partial Fraction decomposition?

Junior Member
Re: Determine if series convergent/divergent. Find its sum
2/(n^2  1)
A/(n + 1) + B/(n  1)
2 = An  A + Bn + B
2 = n(A + B) + (B  A)
B  A = 2
A + B = 0
2B = 2, B = 1, A = 1
so:
1/(n + 1) + 1/(n  1)
Bringing the summation index to n=1, I get:
1/(n + 2) + 1/n
a1 = 1/3 + 1
a2 = 1/4 + 1/2
a3 = 1/5 + 1/3
a4 = 1/6 + 1/4
a5 = 1/7 + 1/5
Ok, now I see it is a telescoping series. I'm kind of unsure how to get the summation though. I can clearly see that 1 + 1/2 will be part of it. I'm not sure about the part involving the n's though. How do I determine that? ie: the answer in this problem is:
1 + 1/2  1/n  1/(n + 2). I'm not sure how to determine the 1/n  1/(n + 2) bit in these types of problems?

Junior Member
Re: Determine if series convergent/divergent. Find its sum
Your summation from n=2 to infinity of: 2/(n^2  1)
The wellknown p=series is
. . .sum 1/(n^p)
which converges for p > 1.
Use the LIMIT COMPARISON TEST:
If lim( a[n]/b[n]) = c > 0, then a,b converge or diverge together.
Now 1/n^2 has p = 2, and converges.
. . .[ 2/(n^2  1) ] / [ 1/n^2 ]
. . .= [ 2n^2 ] / [ n^2  1 ]
. . .= 2 / [1  (1/n^2)]
...which goes to 2, as n goes to infinity. So both series converge. Now for the sum.
Yes, you can do the P.F. decomp: You got:
. . .sum [ 1/(N + 1) + 1/(N  1)]
Write some terms at the beginning and the 'end'. Go from n = 2 to n = k:
. . .A[2] = 1/3 + 1/1
. . .A[3] = 1/4 + 1/2
. . .A[4] = 1/5 + 1/3
. . .A[5] = 1/6 + 1/4
. . .A[k3] = 1/(k  2) + 1/(k  4)
. . .A[k2] = 1/(k  1) + 1/(k  3)
. . .A[k1] = 1/k + 1/(k  2)
. . .A[k] = 1/(k + 1) + 1/(k  1)
Now look at the sum from n = 2 to k: All the fractions disappear EXCEPT one from each of the last two terms:
. . .1/1  1/k  1/(k+1)
Now as k> inf, the second term and third fractions > 0 and the sum > 1.
_____________________________
Edited by stapel  Reason for edit: Fixing formatting.

Junior Member
Re: Determine if series convergent/divergent. Find its sum
I see now how it was arrived at. Thanks
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