Need help with an infinite series

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Discussion Overview

The discussion revolves around an infinite series defined by a sequence of operations involving factorials, with participants exploring its properties, convergence, and potential methods for manipulation. The scope includes mathematical reasoning and exploratory problem-solving related to series convergence and summation techniques.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a series defined by operations on factorials at odd indices and suggests splitting it into two series of positive and negative terms.
  • Another participant calculates the partial sums of the series and concludes that it does not converge, questioning the initial understanding of the problem.
  • A later reply clarifies that the series is not a partial sum but rather a sequence of terms, reiterating the convergence issue.
  • Participants discuss the implications of absolute convergence and express uncertainty about the validity of their approaches.
  • One participant shares a related series involving logarithms and expresses caution about their derivation, indicating a lack of verification.
  • Another participant suggests a more complex formulation involving nested summations and expresses dissatisfaction with the complexity of the expression.

Areas of Agreement / Disagreement

Participants generally do not reach consensus on the convergence of the series or the correct formulation of the problem. Multiple competing views and uncertainties remain throughout the discussion.

Contextual Notes

There are limitations regarding the assumptions made about convergence, the definitions of the series, and the mathematical steps involved in the proposed solutions. Some participants express confusion about the overall structure of the problem.

Who May Find This Useful

Readers interested in infinite series, convergence issues, and advanced mathematical reasoning may find this discussion relevant.

mesa
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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.
 
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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.
 
jbunniii said:
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.

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

jbunniii said:
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...).

jbunniii said:
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
 
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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?
 
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