Need help with an infinite series

[mesa]

I have a series that takes steps of '2' which requires an operation starting from n=1 to do the following,

@n=1 (n-1)!
@n=3 (n-3)!-(n-1)!
@n=5 (n-5)!-(n-3)!+(n-1)!
@n=7 (n-7)!-(n-5)!+(n-3)!-(n-1)!
etc. etc.

Any ideas?

*EDIT*
Come to think of it,
This problem would probably be easier to solve if we split this thing up into two infinite series where one only has positive terms and the other all the negatives and then put them back together after the fact. I tried this with something similar not too long ago and it ended with a solution.

 

[jbunniii]

So the partial sums of your series are

$0!$
$0! - 2!$
$0! - 2! + 4!$
$0! - 2! + 4! - 6!$

etc.? In other words:
$$\sum_{n=0}^{\infty} (-1)^n (2n)!$$
Clearly this series does not converge, because the terms $(-1)^n (2n)!$ do not converge to zero. Maybe I'm misunderstanding the problem.

Also, in general, your proposal to split the series into positive and negative terms will only work if the series is ABSOLUTELY convergent.

 

[mesa]

So the partial sums of your series are

$0!$
$0! - 2!$
$0! - 2! + 4!$
$0! - 2! + 4! - 6!$

etc.? In other words:
$$\sum_{n=0}^{\infty} (-1)^n (2n)!$$
Clearly this series does not converge, because the terms $(-1)^n (2n)!$ do not converge to zero. Maybe I'm misunderstanding the problem.

That is a wonderful solution but I should have specified this isn't a partial sum, that is what each consecutive term needs to be.

Also, in general, your proposal to split the series into positive and negative terms will only work if the series is ABSOLUTELY convergent.

This is only one 'part' of the solution, if you would like to see the remainder I could post it. As far as the technique, I used exactly the same method when deriving this guy,
$$ln(2)=\sum_{n=2}^{\infty} (-1)^n ((n!+n(-1)^n ))/(n+1)!$$
Although I should be careful about boasting, I still haven't checked this one outside of going over the derivation...

 

[jbunniii]

That is a wonderful solution but I should have specified this isn't a partial sum, that is what each consecutive term needs to be.

So is this what you want:
$$\sum_{m=0}^{\infty}\left(\sum_{n=0}^{m} (-1)^n (2n)!\right)$$
This doesn't converge either, because the inner series doesn't converge at all, let alone to zero.

This is only one 'part' of the solution, if you would like to see the remainder I could post it. As far as the technique, I used exactly the same method when deriving this guy,
$$ln(2)=\sum_{n=2}^{\infty} (-1)^n ((n!+n(-1)^n ))/(n+1)!$$
Although I should be careful about boasting, I still haven't checked this one outside of going over the derivation...

I think it would be helpful if you could post the full problem, because I think I'm still misunderstanding your question.

 

[mesa]

So is this what you want:
$$\sum_{m=0}^{\infty}\left(\sum_{n=0}^{m} (-1)^n (2n)!\right)$$

Exactly!
But I want it all under one summation.

This doesn't converge either, because the inner series doesn't converge at all, let alone to zero.

Yeah, I am not surprised. This is only a portion of the numerator for the series so convergence doesn't matter (more on this below...).

I think it would be helpful if you could post the full problem, because I think I'm still misunderstanding your question.

I just happened to come across an 'opening' for an infinite series and decided to go for it. Here is how it needs to work,$$Pi=\sum_{n=1}^{\infty} i^(n(n+3)) 4n(what we are working on)/(n+1)!$$

That 'i' part is not coming up correctly but it is supposed to be i^(n(n+3)) so the series will go ++--++--++--...

***EDIT*** Sorry, that is not completely correct, I forgot I already broke this piece out of a larger problem. The series should go (what we are working on) on two consecutive steps. In other words we would have,

@n=1 0!
@n=2 0!
@n=3 0!-2!
@n=4 0!-2!
@n=5 0!-2!+4!
@n=6 0!-2!+4!
etc. etc.

Sorry about that jbunnii, sometimes I focus so much on one part the others get temporarily lost :P

 

[mesa]

Well, here is what I have,

$$Pi=\sum_{n=1}^{\infty} [(8n-4)/(2n)! \sum_{m=0}^{n-1} (2m)!(-1)^m+(8n)/(2n+1)! \sum_{m=0}^{n-1} (2m)!(-1)^m]$$

If I did the algebra correctly then this should be correct. I am not too happy about the 'summations within the summation' part but I haven't given up on that yet...

Aside from the equation suffering from a mild form of 'Rube Goldbergianism' what do you guys think?