#1

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.

∑ 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,*

#2

I'm sorry I can't read.

#3

jokes i should send u to my brother he is sick at it. sadly hes prob getting pissed

#4

Brutal.

#5

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.

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,*

#6

...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...

The other way I can see of doing it is partial fractions but that's not calculus...

#7

Go look in the pit. Does this look like a place where intelligent people go?

I suggest a math forum or something.

I suggest a math forum or something.

#8

∑ 2/[(n^2)-1) = 3/2

Take that Society!

Take that Society!

#9

its 2/3

=2/2^2-1

=2/4-1

=2/3

=2/2^2-1

=2/4-1

=2/3

#10

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

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,*

#11

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.

#12

I dont understand why you'd use fractions

#13

as in

but there is rules when you have nothing above / below it

but there is rules when you have nothing above / below it

#14

ohh are you trying to find the limit?

lim n->2 ??

lim n->2 ??

#15

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

#16

Well, wish I could help, but I'm only starting differential equations in Sept.

#17

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.

∑ 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,*

#18

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.

#19

Any equations in the textbook? I'm sure there was one last time I remembered multi-variable cal.

#20

_______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

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

#21

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.

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.

#22

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

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,*