A tough sum:

bobbym · 2012-06-29 07:52:30

The above sum appeared in another thread and is solved by somewhat mysterious trigonometric products. Let's see if the methods of experimental math can shed some light on that.

If we proceed experimentally then we would try to sum the above to as many digits as possible. This is not mere numerics as some people call it. If you have been following along then you know that we often can get a closed form just by those "mere numerics."

Our first problem is the slow convergence of S. We first change S into the more convenient sum,

we start to play.

It is clear that convergence is very slow and we would need zillions of terms to get many digits. Enter the great Kummer, father of the so called "Kummer Transformation!"

Supposing we subtract a known series from S. We will use the zeta functions for this purpose.

What have we accomplished? Look at the RHS we see a sum that has a dominating term of n^4 in the denominator. S originally had a dominating term of n^2 in the denominator. 1 / n^4 converges faster than 1 / n^2. Why? The terms shrink faster.

Here is the good part. We can do it again!

Look at the RHS. It has a dominating term of n^6 which should converge faster than the last one of n^4.

Again? Why not!

A dominating term of n^8. We continue to do this as long as we like. Normally we would stop at some convenient value which makes convergence quick and then PSLQ the answer. Here we have the possibility of continuing the process an infinite number of times. We get:

The sum on the right can replaced by 1 / 2 which can be verified numerically. This is as rigorous as I can get with that right now. So we have:

We rearrange this to,

We can drop the 1 / 2 because the first term in S starts at 2.

What have we done? We have replaced S with an infinite sum of zeta functions. Why is this better? Well, we can use a generating function to sum it!

If you know your gf's then you will recognize a series that is close to 6). It is Coth(x). The Mclaurin series for it is,

we can now use the shift operator on the series by multiplying by x.

Multiply both sides by 1 / 2.

Subtract 1 / 2 from both sides.

Substitute x = π.

The RHS of 7) is the RHS of 6)! so

anonimnystefy · 2012-06-29 08:06:47

A cool way! But how do you know that the terms of pi*coth(pi)/2 are the same as the terms in the sum?

bobbym · 2012-06-29 08:11:11

You mean 6) and 7)?

anonimnystefy · 2012-06-29 08:24:42

Everything up to 7), inclusive, is okay. But I am not sure how we can be certain that what we got by manipulating the coth series is the same as S.

Last edited by anonimnystefy (2012-06-29 08:25:14)

bobbym · 2012-06-29 08:31:28

The RHS of 7) is the RHS of 6)! so

I am not understanding you. Is this the line that you mean?

anonimnystefy · 2012-06-29 08:39:15

Exactly.

bobbym · 2012-06-29 08:41:44

Since the time of Euler and the Basel problem all the even zeta functions have known closed forms. Take every sum in 6) and do it and you will see them match every term in 7).

anonimnystefy · 2012-06-29 08:44:07

Have closed forms been found for odd zeta functions?

bobbym · 2012-06-29 08:45:15

Do you understand that 6) and 7) are the same.

Have closed forms been found for odd zeta functions?

No closed forms are known for them.

Math Is Fun Forum

#1 2012-06-29 07:52:30

A tough sum:

#2 2012-06-29 08:06:47

Re: A tough sum:

#3 2012-06-29 08:11:11

Re: A tough sum:

#4 2012-06-29 08:24:42

Re: A tough sum:

#5 2012-06-29 08:31:28

Re: A tough sum:

#6 2012-06-29 08:39:15

Re: A tough sum:

#7 2012-06-29 08:41:44

Re: A tough sum:

#8 2012-06-29 08:44:07

Re: A tough sum:

#9 2012-06-29 08:45:15

Re: A tough sum:

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