∑i=1 n(-1 )i (2i+1 )3 /((2i+1 )4 +4)
By Tania Perez on Friday, September 07, 2001 - 11:31 pm:

How could I calculate the series below as a function of n?... Thanks in advance, Tania

∑i=0 n(-1 )i (2i+1 )3 /((2i+1 )4 +4)
By Arun Iyer on Sunday, September 09, 2001 - 08:42 pm: the above given expression can be rewritten as...

∑i=0 n(4i+1 )3 /[(4i+1 )4 +4]- ∑i=0 n(4i+3 )3 /[(4i+3 )4 +4] i am though quite stuck here.Hope someone proceeds from hereonwards.

love arun

By David Loeffler on Monday, September 10, 2001 - 04:34 pm:


I very much doubt it can be done for general n, unless the expression splits up into some diabolically neat partial fractions expression. We can factorise the denominator - (2i+1)4 + 4 = (2i+1)4 + 4(2i+1)2 + 4 - 4(2i+1)2
=( (2i+1)2 + 2)2 - (2(2i+1))2 =
( (2i+1)2 + 2(2i+1) + 2) ( (2i+1)2 - 2(2i+1) + 2), but goodness knows how that would help. This is, of course, (4i2 + 1)(4(i+1)2 +1); in fact, maybe one could convert it into partial fractions. I will have a look when I get home (my boss isn't dead keen on me reading this board during working hours).

David

By David Loeffler on Monday, September 10, 2001 - 04:47 pm:

I think it's actually (-1)n (n+1)/(4n2 +8n+5). I calculated the first six terms and it agrees.

Looks like it should be easy enough to prove by induction.

David

By Tania Perez on Monday, September 10, 2001 - 06:23 pm:

David, may I know how you get this result?... Tania

By Arun Iyer on Monday, September 10, 2001 - 06:44 pm:

Yeah!!That's quite a neat answer there!!
How did you get it??

love arun

By Tania Perez on Monday, September 10, 2001 - 09:57 pm:

Well, I do the prepositions below but I could not find the result, could you help me?

the denominator can be factored (as David notice)
(2i+1)4 +4 = (2i+1)4 +4(2i+1)2 +4-4(2i+1)2

= ((2i+1)2 +2+2(2i+1))((2i+1)2 +2-2(2i+1))

= (4i2 +8i+5)(4i2 +1)

= (4(i+1)2 +1)(4i2 +1)


using partial fractions, it can be written as
(2i+1)3 /((2i+1)4 +4) = i/(4i2 +1)+(i+1)/(4(i+1)2 +1)
= g(i) + g(i+1)


where
g(i) = i/(4i2 +1)

now the sum can be written in the form
S= ∑i=0 n(-1 )i (2i+1 )3 /((2i+1 )4 +4)

= ∑i=0 n(-1 )i (g[i]+g[i+1])

= ∑i=0 n(-1 )i g(i)-(-1 )i+1 g(i+1)

= ∑i=0 nf(i)-f(i+1)
where
f(i)=(-1)i g(i)

... now, what do you suggest to do?

By David Loeffler on Tuesday, September 11, 2001 - 09:19 am:
You've got it there. In ∑i=0 nf(i)-f(i+1), the whole sum telescopes to f(0)-f(n+1): (f(0)-f(1)) + (f(1)-f(2)) ... + (f(n)-f(n+1)) ...
and all the intermediate terms drop out. Clearly f(0)=0, and f(n+1) = (-1)n+1 (n+1)/(4n2 +8n+5); so f(0)-f(n+1) = (-1)n (n+1)/(4n2 +8n+5) is the sum of the series.

As to how I found it, I calculated the first four or five terms and guessed! The numerators were obvious and it wasn't difficult to fit a quadratic to the denominators (do you know how to do this - to fit a polynomial to a sequence where you know the first few terms?). So my solution wasn't nearly as elegant as yours.

David
By Arun Iyer on Tuesday, September 11, 2001 - 07:15 pm:

Nice solution!!
good start from Tania and a good end from David.

love arun