hello, pit. i am stumped by this Calc problem on infinite series
∑ 2/[(n^2)-1)
n=2

I know the answer is 3/2 and it's telescoping but i just cant seem to set it up right and get the answer on my own.
Last edited by Burdell at Jul 3, 2008,

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I think you meant 2/3, use google to look it up. There are alot of math sites out there.

I dropped out of calc, but 2/3 would be considered an infintecimal number (.666 repeated) and woul be unresolved.
Last edited by _Ox_ at Jul 3, 2008,
...Don't you just stick the "n=2" into the formula?

The other way I can see of doing it is partial fractions but that's not calculus...
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Go look in the pit. Does this look like a place where intelligent people go?

I suggest a math forum or something.
∑ 2/[(n^2)-1) = 3/2

Take that Society!
its 2/3

=2/2^2-1
=2/4-1
=2/3
the back of the book says 3/2.

and you definitely use partial fractions and get (1/x-1)-(1/x+1). but the problem i am having is the once i cancel the repeated terms, they give me this fraction thats way off from 3/2. i just wanted to see how to see the problem is set up and what terms cancel and stuff.

oh well. its a lost cause apparently. ive seen some people answer calc problems on here before so id thought id give it a stab.

Edit:
^No, its a series. meaning you start with n=2 and then increase n until infinity, adding every value. the sigma looks like this (idk if it will show up right):

n=2
Last edited by Burdell at Jul 3, 2008,
the ∑ symbol means you do more then just stick in the number. I can't remember it has been to many years. Look up what that symbol means.
I dont understand why you'd use fractions
as in

but there is rules when you have nothing above / below it
ohh are you trying to find the limit?

lim n->2 ??

The subscript gives the symbol for an index variable, i. Here, i represents the index of summation; m is the lower bound of summation, and n is the upper bound of summation. Here i = m under the summation symbol means that the index i starts out equal to m. Successive values of i are found by adding 1 to the previous value of i, stopping when i = n. We could as well have used k instead of i, as in
Well, wish I could help, but I'm only starting differential equations in Sept.
6
∑ 2/[(n^2)-1)
n=2

2/3 + 2/8 + 2/15 + 2/24
16/24 + 6/24 + 2/15 + 2/24
26/24 + 2/15
13/12 + 2/15

219/180 = 1.216
(answer you gave would be 1.5)

are you sure you are giving us the entire problem? (just throwing in 6 over sigma for kicks)

PS: nm lol i finally read your entire problem (as it goes to infinity it goes to 1.5). As for setting it up so it shows that i don't remember how. Starts at n=2 because we all know what happens if you start at n=1 2/0.
Last edited by ViperScale at Jul 3, 2008,
Quote by ViperScale
6
∑ 2/[(n^2)-1)
n=2

2/3 + 2/8 + 2/15 + 2/24
16/24 + 6/24 + 2/15 + 2/24
26/24 + 2/15
13/12 + 2/15

219/180 = 1.216
(answer you gave would be 1.5)

are you sure you are giving us the entire problem? (just throwing in 6 over sigma for kicks)

its an infinite series problem...there is an ∞ over the sigma.
meh, ive done enough problems for today. ill just try again tomorrow.
Any equations in the textbook? I'm sure there was one last time I remembered multi-variable cal.
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_______x
lim_____∑ 2/[(n^2)-1)
x->inf__ n=2

can't seem to find what i am looking for to help you but best i can say is it should be
http://en.wikipedia.org/wiki/Geometric_progression

Try this, under infinite geometric progression. It's obvious that as x goes to infinite, the result goes to 0, since the power of n on the numerator (0) is less than the denominator.

Quote by keiron_d
thank you sooooooo much for the advice Fast_Fingers...i would hug you if i could...i looooove you!

True love exists in UG. Can you feel it?

Recording Guitar Amps 101
Something close to this, messing something up... hmmm

inf
∑ 2/[(n^2)-1)]
n=2

stick 2 into the forumal to get 2/3 + the lim of n going to inf. (inf n = N)

2/3 + 2/[(N^2)-1)]

take out 2 from it so you have ∑ 2/3 + 2/[(N^2)-1)] then solve

inf^2 = inf
inf - 1 = inf
1/inf = 0

2/3 + 0 =

http://en.wikipedia.org/wiki/Telescoping_series
Last edited by ViperScale at Jul 3, 2008,