å
1/n2k



By Andrew Hodges on Monday, August 06, 2001 - 03:43 pm:

what is the proof showing that:


1/1+1/4+1/9+1/16+¼ = p2/6?

Can that method of proof be used to determine the sum of any series like the above, with a higher (even) power, e.g.

1/1 + 1/16 + 1/81..........
By Arun Iyer on Tuesday, August 07, 2001 - 07:51 pm:

Andrew,
for 1,
you will find an answer to this here

For your second question,
these series form a special class of series arising from the Riemann Zeta function, though the proof of the above question may not be as useful.

love arun


By Brad Rodgers on Wednesday, August 08, 2001 - 05:15 am:

For the question about

1+1/16+1/81+...


known as z(4), I think Euler proved that this is equal to p4/90 using Bernoulli numbers, I'll try to find a page on this.
Brad
By Andrew Hodges on Wednesday, August 08, 2001 - 12:14 pm:

Can anybody notice any pattern at all in this sequence?

6, 90, 945, 9450, 93555

Andrew


By Brad Rodgers on Friday, August 10, 2001 - 09:45 pm:

The pattern involves Bernoulli numbers, and information about Bernoulli numbers and the pattern (near the bottom) can be found at this site


(Incidentally, the pattern is the denominators of z(2k), the numerators being p2k)
I'm not sure that the Bernoulli numbers could be expressed as some polynomial, and if so, I have no idea how, but the website gives a couple of ways to find Bernoulli numbers.

I'm still working on a proof for the relation given between the Zeta function and Bernoulli numbers.

Brad
By Arun Iyer on Friday, August 10, 2001 - 09:57 pm:

Ah!! Interesting..

So now you can write the given pattern to be...
2(2k)!/4k |B2k |

Am i right??
love arun


By Andrew Hodges on Saturday, August 11, 2001 - 08:07 pm:
Yes, I thought they were the denomiators of z(2k). I calculated them empirically on my calculator. Bernoulli numbers sound interesting. I'll take a closer look at those thanks!