Challenge Problem:Find

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- January 23rd 2011, 04:57 AMTheCoffeeMachineAn Alternating Infinite Series
**Challenge Problem:**Find

Moderator approved CB - January 23rd 2011, 07:16 AMalexmahone
- January 23rd 2011, 10:45 AMchisigma
Let suppose to have the periodic function defined in as...

(1)

... so that it can be written in Fourier series as...

(2)

The can be computed in the usual way...

(3)

Now we combine the results and obtain ...

(4)

Kind regards

- January 23rd 2011, 03:26 PMProve It
- January 23rd 2011, 05:36 PMTheCoffeeMachine
I don't think it telescopes to any simple value or a familiar series...

On the right track, though. I did exactly the same up to this step:

I think what you've found is the sum of the series , not

EDIT: I'm wondering whether can be found in the same way? It would be nice! - January 23rd 2011, 10:45 PMalexmahone
- January 23rd 2011, 11:14 PMsimplependulum
First resolve the summand into partial fractions and we can see that it is equal to ( the result first obtained by alex )

or

Sub. we have

- January 23rd 2011, 11:37 PMTheCoffeeMachine
- January 24th 2011, 12:11 AMTheCoffeeMachine
The sum that Alex and Chisigma calculated earlier can also be found in the same way:

And this integral is rather much simpler than the other one:

- January 26th 2011, 05:04 AMTheCoffeeMachine
It turns out that is also a special case of the more general result:

which is very handy and punishes lots of sums, but I don't seem to be able to prove it! - January 26th 2011, 12:45 PMKrizalid
an alternate computation could be the following:

the last integral is routine, and yields the expected result.