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# Sum (r(r^2-1))^-1 watch

1. I'm trying to sum this series from 2 to infinity, I took it into partial fractions:

= 1/r-1 - 1/r+1 - 2/r

I'm trying to sum this, and I can see that nearly all the terms will cancel, apart from

1 + 1/2 - sum of 2/r from 2 to infinity

have I made a mistake? surely that last term can't be summed, and if it's approximated using 1/2 + ln N, then you just end up with the sum = -infinity, which is obviously nonsense. What am I doing wrong? lex
2. Summation using the method of differences (which it looks like you're using) only works for finite series.
3. You have to be careful

1/ [ r (r-1) (r+1) ] = 1/2 [ 1/(r-1) - 1/r ] + 1/2[ 1/(r+1) - 1/r ]

Each of the 2 terms on the RHS can be summed easily ( try writing out the first 10 terms or so ).

Giving sum of 1/2[ 1 - 1/2 ] = 1/4

I think.
4. (Original post by geebee2)
Giving sum of 1/2[ 1 - 1/2 ] = 1/4

I think.
I get 1/4 too.

The problem is a disguised version of

(sum over r >= 1) 1/r(r + 1)(r + 2).

Since everyone knows how to calculate

(sum over r >= 1) 1/r(r + 1)

and noone knows how to calculate

(sum over r >= 1) 1/r

(because it doesn't exist), the obvious next question is:

(sum over r >= 1) 1/r(r + 1)(r + 2)(r + 3).

Rep for the first correct answer!
5. 1/18?
6. (Original post by fishpaste)
1/18?
Arrgh - 9 minutes too late

S = 1/r(r+1)(r+2)(r+3) = (1/6){1/r - 3/(r+1) + 3/(r+2) - 1/(r+3)}
S = (1/6){1/r - 1/(r+1) - [2/(r+1) - 2/(r+2)] +1/(r+2) - 1/(r+3)}

Summing from r to n gives,

S = (1/6){1 - 1/(n+1) - 2[1/2 - 1/(n+2)] + 1/3 - 1/(n+3)}

as n -> infty

S -> (1/6){1 - 2/2 + 1/3)
S -> 1/18
======
7. (Original post by Fermat)
Arrgh - 9 minutes too late

S = 1/r(r+1)(r+2)(r+3) = (1/6){1/r - 3/(r+1) + 3/(r+2) - 1/(r+3)}
S = (1/6){1/r - 1/(r+1) - [2/(r+1) - 2/(r+2)] +1/(r+2) - 1/(r+3)}

Summing from r to n gives,

S = (1/6){1 - 1/(n+1) - 2[1/2 - 1/(n+2)] + 1/3 - 1/(n+3)}

as n -> infty

S -> (1/6){1 - 2/2 + 1/3)
S -> 1/18
======
what module is this.
p1 pr p2
8. (Original post by mathematician)
what module is this.
p1 pr p2
(r(r^2-1))^-1
this rings a bell from p4 though?
9. Ehh cheers.

Rep for the first person to get

1/r(r+1)(r+2)(r+3)(r+4)(r+5)

*has written a quick script to calculate them*
10. (Original post by fishpaste)
Ehh cheers.

Rep for the first person to get

1/r(r+1)(r+2)(r+3)(r+4)(r+5)(r+6)( r+7)

*has written a quick script to calculate them*
is this a proper question?
11. (Original post by S1M)
is this a proper question?
I edited and shortened it a bit from the post you quoted, but yes, I know the answer if you want to try.
12. (Original post by fishpaste)
I edited and shortened it a bit from the post you quoted, but yes, I know the answer if you want to try.
well IMO it looks like a random long winded question... i got 2 exams to prepare for still... so if nobodies given it a shot by tuesday... ill go for it... doesnt look too difficult... just looks very long winded... its just p4 right?
13. (Original post by S1M)
well IMO it looks like a random long winded question... i got 2 exams to prepare for still... so if nobodies given it a shot by tuesday... ill go for it... doesnt look too difficult... just looks very long winded... its just p4 right?

Yeah same method as before, but you'll find yourself writing out tonnes of beginning and end of the series just to work out what goes I guess, and so it might not be the most rewarding thing to do.
14. (Original post by fishpaste)
Ehh cheers.

Rep for the first person to get

1/r(r+1)(r+2)(r+3)(r+4)(r+5)

*has written a quick script to calculate them*
0.001615161792...?
15. (Original post by fishpaste)
Ehh cheers.

Rep for the first person to get

1/r(r+1)(r+2)(r+3)(r+4)(r+5)

*has written a quick script to calculate them*
Sum = 1/600

*has written a quick program in C to calculate them*
16. rotf did we all get our computers to do it?

Here's mine: http://homepage.ntlworld.com/jamiek/quack.html

Allow the script to run for half a min or so if it asks you if you'd like to abort.
17. 1/[r(r + 1)(r + 2)(r + 3)(r + 4)]
-
1/[(r + 1)(r + 2)(r + 3)(r + 4)(r + 5)]
=
5 / [r(r + 1)(r + 2)(r + 3)(r + 4)(r + 5)]

Sum over r >=1:

1/5! = 5 (sum over r >= 1) 1 / [r(r + 1)(r + 2)(r + 3)(r + 4)(r + 5)].

So the sum on the rhs equals 1/(5*5!) = 1/600.

Now try:

(sum over r >= 1) 1 / [r(r + 1)(r + 3)(r + 4)].
18. (Original post by fishpaste)
rotf did we all get our computers to do it?

Here's mine: http://homepage.ntlworld.com/jamiek/quack.html

Allow the script to run for half a min or so if it asks you if you'd like to abort.
Well, about 2 seconds after it started to run, I got an error message to the effect that IE was running slow and my computer might become unresponsive, so I aborted it!!

Here's my code:
Attached Files
19. SUM.doc (268 Bytes, 81 views)
20. (Original post by fishpaste)
Ehh cheers.

Rep for the first person to get

1/r(r+1)(r+2)(r+3)(r+4)(r+5)

*has written a quick script to calculate them*
Sum between 1 and n is:

1/600-1/[120(n+1)]+1/[30(n+2)]-1/[20(n+3)]+1/[30(n+4)]-1/[120(n+5)]

(done by splitting into partial fractions then crossing out. Far too long to type out here )

So sum to infinity would be 1/600.
21. To generalise the result try: Prove that for every integer k, with k>1

(sum over r >= 1) 1 / [r(r + 1)(r + 2)...(r + k)] = 1/(k!k)

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