Need help with an infinite series

In summary, the conversation is discussing a series that requires an operation starting from n=1 and has steps of '2'. The proposed solution is to split the series into two infinite series, one with positive terms and the other with negative terms, and then combine them after the fact. However, it is noted that this will only work if the series is absolutely convergent. The full problem involves finding the value of Pi using an infinite series, where the series alternates between positive and negative terms on consecutive steps. One possible solution involves breaking the problem into two parts and using algebra to simplify the equation.
  • #1
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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  • #2
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.
 
  • #3
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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  • #4
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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  • #5
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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  • #6
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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1. What is an infinite series?

An infinite series is a mathematical concept that involves adding an infinite number of terms together. Each term in the series is typically related to the previous term by a specific pattern or rule.

2. How do you know if an infinite series converges or diverges?

There are several tests that can be used to determine if an infinite series converges or diverges. Some common tests include the ratio test, the comparison test, and the integral test. It is important to note that not all infinite series will have a definite answer and may require more advanced techniques to determine convergence or divergence.

3. Can you give an example of a convergent infinite series?

One example of a convergent infinite series is the geometric series, where each term is a constant multiple of the previous term. For example, the series 1/2 + 1/4 + 1/8 + 1/16 + ... converges to a value of 1.

4. How can infinite series be used in real life?

Infinite series are used in many areas of science and engineering, such as in calculating the sum of an infinite number of measurements or in solving differential equations. They are also used in computer graphics to create smooth curves and animations.

5. What are some common mistakes when working with infinite series?

One common mistake is assuming that a series will always converge just because the terms are getting smaller. It is important to use appropriate tests to determine convergence. Another mistake is forgetting to consider the starting value of the series, as it can affect the final value of the sum.

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