Post Reply 
Series test for convergence/divergence
05-09-2018, 11:15 AM
Post: #1
Series test for convergence/divergence
How could the hp prime be used to directly determine convergence/divergence for a series, (not evaluate to the sum!)?

Example (converge/diverge)?:

∑ [n=1,∞] (2*n^2 + n) / √(4*n^7 + 3);

The idea is to use the various convergence tests, (such as done by Wolfram Alpha), for this example.

-Dale-
Find all posts by this user
Quote this message in a reply
05-09-2018, 12:33 PM
Post: #2
RE: Series test for convergence/divergence
Run series((2*n^2 + n) / √(4*n^7 + 3),n=inf) and look at the series expansion main term, here it's equivalent to n^(-3/2) (const sign) hence convergent.
Find all posts by this user
Quote this message in a reply
05-10-2018, 08:25 AM
Post: #3
RE: Series test for convergence/divergence
Thank you Parisse. While the Series expansion can help, a command which returns a result with direct indication of convergence/divergence, and possibly the confirming tests used to obtain the result, would be a nice enhancement.

Wolfram Alpha has a nice presentation for this, as shown in the attachment. Something like this would be very useful for students studying this subject.

   
Find all posts by this user
Quote this message in a reply
05-10-2018, 01:31 PM
Post: #4
RE: Series test for convergence/divergence
I don't find the wolfram alpha text very informative "By comparison test, the series converge", it's much more informative to tell that the equivalent is n^(-3/2).
Yes, it could be a nice addition, like for example tabvar.
Find all posts by this user
Quote this message in a reply
05-10-2018, 02:09 PM
Post: #5
RE: Series test for convergence/divergence
I find Wolfram's technique to be very informative. This may be an American vs non-American teaching difference, though. During a study of series testing for convergence/divergence, a variety test methods are taught, with problem examples provided to be solved outside of a classroom. Wolfram is showing that some tests are inconclusive, where others may reveal convergence or not. Some problems can be mysterious, and information, (like Wolfram provides), can be useful for making progress, at such times.

I suppose a user program for this could be made, but since we have so many commands, that fill so many mathematical needs, that this one might also be included. I read, (in an older post, somewhere), that you were thinking of this for a giac addition. It would be very nice, if included in the prime!
Find all posts by this user
Quote this message in a reply
05-11-2018, 03:30 PM
Post: #6
RE: Series test for convergence/divergence
They just give some hints, but no proof. I mean, no math teacher would accept "the series is convergent by comparison" without details.
Yes, it would be nice to add that, but this is low priority currently.
Find all posts by this user
Quote this message in a reply
05-11-2018, 04:17 PM
Post: #7
RE: Series test for convergence/divergence
(05-11-2018 03:30 PM)parisse Wrote:  They just give some hints, but no proof. I mean, no math teacher would accept "the series is convergent by comparison" without details.

During the topic under discussion, a proof is given by the instructor. After the proof has been provided, it is expected that those methods will be used for subsequent problem solving; and once an outcome has been reached a result is generally shown as: "convergence, (or divergence), for ∑ bn is demonstrated, therefore ∑ an is also convergent, (or divergent), {by the relevant series test}." So the "details" ARE part of a student's solution.

The calculator would be useful in validating intermediate steps, and if the calculator's solution was of a form similar to Wolfram's, it would be even more valuable to the student, and the instructor as well, since the student's answer should follow the (hopefully, correct) calculator results.
Find all posts by this user
Quote this message in a reply
05-11-2018, 04:25 PM
Post: #8
RE: Series test for convergence/divergence
Hi Parisse,

I see that the value of the main term is n^(-3/2) but I don't know the connection between that and whether the series converges or diverges. I guess what I'm asking is, if I apply the series command to some function, what am I looking for in the output to see if the series converges or diverges? I tried some different series that I know are convergent and divergent, but I couldn't see a pattern in the output that helped me connect the two.

Thank you for your time.
Find all posts by this user
Quote this message in a reply
05-11-2018, 06:28 PM
Post: #9
RE: Series test for convergence/divergence
(05-11-2018 04:25 PM)Benjer Wrote:  Hi Parisse,

I see that the value of the main term is n^(-3/2) but I don't know the connection between that and whether the series converges or diverges. I guess what I'm asking is, if I apply the series command to some function, what am I looking for in the output to see if the series converges or diverges? I tried some different series that I know are convergent and divergent, but I couldn't see a pattern in the output that helped me connect the two.

Thank you for your time.
n^(-3/2)==1/n^(3/2) therefore as n gets larger
OR
limit(n^(-3/2),n,∞)
Find all posts by this user
Quote this message in a reply
05-11-2018, 06:56 PM
Post: #10
RE: Series test for convergence/divergence
sum(1/n^alpha,n) is convergent if alpha>1, divergent for alpha<=1 (this is easy to prove by comparing with int(1/x^alpha,x)). If f(n) is equivalent to g(n) and g(n) has constant sign then sum(f(n),n) and sum(g(n),n) are both convergent or both divergent.
It's more complicated for non constant sign term, like sum((-1)^n/n^alpha,n) is convergent for alpha>0, but sum((-1)^n/f(n),n) may be divergent even if f(n) is equivalent to n^alpha, for this kind of series you will usually need a series expansion of f(n) at n=inf until the remainder of the expansion is O(1/n^alpha) with alpha>1. Like for example sum((-1)^n/sqrt(n+(-1)^n*sqrt(n)),n)
Find all posts by this user
Quote this message in a reply
05-11-2018, 11:17 PM
Post: #11
RE: Series test for convergence/divergence
Thank you for the clarification.
Find all posts by this user
Quote this message in a reply
Post Reply 




User(s) browsing this thread: 1 Guest(s)