HP Prime Infinite Sum Bug?

11052014, 07:58 AM
(This post was last modified: 11052014 08:06 AM by kcnicho.)
Post: #1




HP Prime Infinite Sum Bug?
Hi All,
I posted this question over at the HP Calculator site (http://h30499.www3.hp.com/t5/Calculators...p/6662418), thought I'd ask here as well. HP Prime, in CAS mode, when I enter "sum(1/sqrt(n), n, 1, infinity)", the Prime returns a screen full of warnings (see attachments.) I believe the correct answer should be "+ infinity", as I get from Wolfram Alpha as well as the TINspire CAS. Has anyone else seen this? Can anyone else duplicate this behavior on their own machine? I verified this on the Prime Virtual Calculator, Version 2014 3 31, Rev 6031 (latest firmware, as far as I know.) Thanks,  Kevin 

11052014, 12:34 PM
Post: #2




RE: HP Prime Infinite Sum Bug?
Sorry, convergence of series is not yet implemented in giac. I plan to do that as part of the new step by step feature.
sum tries to find a closed form for the discrete antiderivative (that's why you see all these messages), it fails, hence returns the sum unevaluated. 

11052014, 07:17 PM
Post: #3




RE: HP Prime Infinite Sum Bug?
(11052014 12:34 PM)parisse Wrote: part of the new step by step feature.Oh, that's exciting (Coming to KhiCAS too ? ) TIPlanet.org coadministrator 

11062014, 07:05 AM
Post: #4




RE: HP Prime Infinite Sum Bug?
If I can find a way to display intermediate computations on the nspire screen.


11062014, 07:55 AM
Post: #5




RE: HP Prime Infinite Sum Bug?
Thanks for the (definitive) answer!
Happy to hear that things are in the works to address these types of problems in future firmware releases.  Kevin 

01282015, 02:03 PM
Post: #6




RE: HP Prime Infinite Sum Bug?
(11052014 12:34 PM)parisse Wrote: Sorry, convergence of series is not yet implemented in giac. I plan to do that as part of the new step by step feature. I hope you could do it soon in a next firmware upgrade, thank you! It would be very interesting... Salvo 

01282015, 06:32 PM
Post: #7




RE: HP Prime Infinite Sum Bug?
...another Series I would se calculated in Prime:
sum((1)^(k+1)/K^2, k, 1, infinity) It should give π^2/12 Thank you Salvo 

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