Alex Miller Alex
Posted on Sunday, 13 April, 2003 - 12:15 am:
I have been wondering how people devise formulas to use in order to get exact sums of infinite series. The formulas in question are the ones using multiple integrals. For instance, how did they come up with:
picture
Hauke Worpel
Posted on Sunday, 13 April, 2003 - 01:19 am:
For the last sum, I would try breaking it into partial fractions. That kind of sum generally gives you a telescoping sum (where all but a few terms cancel out).
Alex Miller Alex
Posted on Sunday, 13 April, 2003 - 05:53 am:
I tried that, no dice
Julian Pulman
Posted on Sunday, 13 April, 2003 - 11:17 am:

Most of the time, there is no 'generalised method' to always find a sum.
It's quite easy to see why the first two examples give our sums, for example in the 2nd, integrate with respect to x, turn it into a taylor series and integrate once more the result is the left hand side of that equation.
Why is this useful? Well, if we use the substitution x=u-v, y=u+v, we get
LaTeX Image
where S is a square with vertices (0,0),(1,0),(1/2,-1/2),etc.. which we can split in half - after a little playing the answer pi2 /6 comes out moderately easily.

Julian
David Loeffler
Posted on Monday, 14 April, 2003 - 12:09 am:

For these types of series, that decompose into partial fractions without repeated factors, there is a general approach involving the Psi function, but it's too involved to go into here.

However, there is an easy way to make partial fractions work here. Try rearranging what you get into the sum of an alternating harmonic series plus another term.

I get something like 1/(4n (4n-1) (4n-2) (4n-3)) = 1/6 ( -1/4n + 1/(4n-1) - 1/(4n-2) + 1/(4n-3))
+ 1/3 ( 1/(4n-1) - 1/(4n-2) )

Can you see a suitable integral trick for the second part?

David
Alex Miller Alex
Posted on Monday, 14 April, 2003 - 12:53 am:
Please elaborate.
-Alex
Kerwin Hui
Posted on Monday, 14 April, 2003 - 10:07 pm:
Use
LaTeX Image
and interchange summation and integration (justified by dominated convergence) give you something like (up to multiplicative constants)
LaTeX Image
which is easy to evaluate.

Kerwin
Alex Miller Alex
Posted on Monday, 14 April, 2003 - 10:22 pm:
Ohhh, I understand, now I feel dumb
can you explain the Psi function? It sounds interesting.
-Alex
David Loeffler
Posted on Thursday, 17 April, 2003 - 10:56 pm:
Essentially, it runs like this. Let Γ(x) be defined as ∫0 ∞ tx-1 e-t dt, so for positive integers n, Γ(n)=(n-1)! (you might like to check this, by integration by parts and induction). Then define ψ(x)=d/dx(logψ(x)).

From that it's relatively easy to show that ψ(x+1)=ψ(x)+1/x.

Now, suppose we wanted to evaluate ∑n=1 ∞(1/(4n-1)-1/(4n-2))

We can write this as 1/4 ∑n=1 ∞1/(n-1/4)-1/(n-1/2).

Consider the first N terms in this sum. This will telescope, using the above relations, and become

1/4(ψ(N+3/4)-ψ(N+1/2)-ψ(3/4)+ψ(1/2))

Now, the difference ψ(N+3/4)-ψ(N+1/2) will tend to zero (trust me, it's true) and you can do some integrals to find out what the values of the remaining terms are. This will give you the value of your sum.

David

Alex Miller Alex
Posted on Thursday, 17 April, 2003 - 11:11 pm:
ahhhh, genius. That is some cool stuff. I have a book that goes into detail about the gamma function, but I did not know anything about the Psi function. Do you know of any good online resources to learn more about the psi function?
Thanks a ton David.
-Alex
Demetres Christofides
Posted on Friday, 18 April, 2003 - 09:32 am:
You mean d/dx(logΓ(x)).

Demetres

Alex Miller Alex
Posted on Friday, 18 April, 2003 - 05:14 pm:
Could I use just the gamma function if I thought of the problem as:
int
?
-Alex
David Loeffler
Posted on Friday, 18 April, 2003 - 09:47 pm:
Demetres, thanks, that is of course what I meant.

Alex, you could do that, but I don't think it would help very much!

David
Alex Miller Alex
Posted on Friday, 18 April, 2003 - 09:52 pm:
ha ha, yeah, as I have kinda figured out

Thanks,
-Alex Miller