Determine if series convergent/divergent. Find its sum

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MarkSA

Junior Member
Joined
Sep 8, 2007
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243
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?
 
Have you looked at the Partial Fraction decomposition?
 
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?
 
Your summation from n=2 to infinity of: 2/(n^2 - 1)

The well-known 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[k-3] = -1/(k - 2) + 1/(k - 4)

. . .A[k-2] = -1/(k - 1) + 1/(k - 3)

. . .A[k-1] = -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.
 
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