These are chat archives for gxyd/sum_convergence

11th
Oct 2015
Gaurav Dhingra
@gxyd
Oct 11 2015 06:41
I hope that Product Convergence will be useful, also. So i will be making a PR about the .is_convergent() in Product in SymPy. This will be quite simple to make, but since i will also be adding some latex in docs(which is quite new to me). So will that be useful to have is_convergence for Productss as well.?
Kalevi Suominen
@jksuom
Oct 11 2015 10:27
I have been thinking of the connection between products (1+an)\prod (1 + a_n) and sums an\sum a_n. If all the terms ana_n are positive or negative, then the product and the sum both converge or both diverge. (For negative ana_n, the product is considered divergent if is tends to 0.) However, in general it is possible that the sum may converge even if the product is divergent.
Kalevi Suominen
@jksuom
Oct 11 2015 10:37
Consider, for example, the convergent alternating series (1)n/n\sum (-1)^n/\sqrt{n}. Grouping the terms of the product pairwise we get
(11/2n1)(1+1/2n)<(11/2n)(1+1/2n)=11/2n(1 - 1/\sqrt{2n - 1})(1 + 1/\sqrt{2n}) < (1 - 1/\sqrt{2n})(1 + 1/\sqrt{2n}) = 1 - 1/2n
showing that the product is divergent.
Hence, it would probably be best to consider the absolute convergence of the sum. It will give a sufficient but not necessary condition for the convergence of the product.
Gaurav Dhingra
@gxyd
Oct 11 2015 10:52
Sorry, to interrupt, i have not read your two mentioned points. n=2log(1n)\sum_{n=2}^{\infty} \log(\frac{1}{n}) (sum diverges) while n=21n\prod_{n=2}^{\infty} \frac{1}{n} (produc converges)
Gaurav Dhingra
@gxyd
Oct 11 2015 11:09
So i guess i should use solveset to check for values of abs(sequence) are all <1 < 1 then return True.
Kalevi Suominen
@jksuom
Oct 11 2015 11:35
The product 1/n\prod 1/n is usually considered divergent.
It is usually assumed that the terms are nonzero.
Gaurav Dhingra
@gxyd
Oct 11 2015 11:39
To be clear for definition of wikipedia: "The product of positive real numbers n=1an\prod_{n=1}^{\infty} a_{n} converges to a non-zero real number if and only if the sum n=1logan\sum_{n=1}^{\infty} \log{a_{n}} converges."
Kalevi Suominen
@jksuom
Oct 11 2015 11:39
If any one of the terms is zero, all partial products are zero. That is not accepted as convergent. Moreover, the product should be convergent if and only if the product of the inverses is convergent. (So n\prod n is divergent.)
Yes, non-zero is the point.
Perhaps we could 'generalize' the condition of convergence in such a way that a finite number of the terms could be zero. Then only the tail product would decide the convergence.
Hence, it might not be necessary to test abs(term) < 1.
Gaurav Dhingra
@gxyd
Oct 11 2015 11:45

Perhaps we could 'generalize' the condition of convergence in such a way that a finite number of the terms could be zero. Then only the tail product would decide the convergence.

Yes, thanks for the explanation. It seems more clear now. I will do this point then.

Gaurav Dhingra
@gxyd
Oct 11 2015 12:10
This article also gives justification of your point.
Gaurav Dhingra
@gxyd
Oct 11 2015 19:51
This message was deleted
Gaurav Dhingra
@gxyd
Oct 11 2015 20:16

Perhaps we could 'generalize' the condition of convergence in such a way that a finite number of the terms could be zero. Then only the tail product would decide the convergence.

Can you please give an example of this? I will be more clear.