[seqfan] Re: help need with some Ramanujan identities
Neil Sloane
njasloane at gmail.com
Sat Sep 1 21:27:23 CEST 2018
Added Ram. ref. to A222068, recycled A318590.
Nice work, everyone!
Best regards
Neil
Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com
On Fri, Aug 31, 2018 at 4:03 PM Hugo Pfoertner <yae9911 at gmail.com> wrote:
> Neil, Robert,
> I've edited A222068 accordingly. Robert, could you please insert your sum
> formulas there? A318590 can then be recycled.
> Best regards
> Hugo
>
> On Fri, Aug 31, 2018 at 9:16 PM Neil Sloane <njasloane at gmail.com> wrote:
>
> > In that case (thanks, Robert) they should definitely be merged
> > Best regards
> > Neil
> >
> > Neil J. A. Sloane, President, OEIS Foundation.
> > 11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
> > Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
> > Phone: 732 828 6098; home page: http://NeilSloane.com
> > Email: njasloane at gmail.com
> >
> >
> >
> > On Fri, Aug 31, 2018 at 1:55 PM <israel at math.ubc.ca> wrote:
> >
> > > Modulo questions about rearrangement of non-absolutely convergent
> series,
> > > the result should indeed be (Pi/4)^2.
> > >
> > > Sum_{k>=0} (-1)^k d(2k+1)/(2k+1) = Sum_{k>=0} Sum_{2i+1 | 2k+1}
> > > (-1)^k/(2k+1) (letting 2k+1=(2i+1)(2j+1): note that k == i+j mod 2)
> > > = Sum_{i>=0} Sum_{j>=0} (-1)^(i+j)/((2i+1)(2j+1))
> > > = (Sum_{i>=0} (-1)^i/(2i+1))^2 = (Pi/4)^2
> > >
> > > Cheers,
> > > Robert
> > >
> > > On Aug 31 2018, Hugo Pfoertner wrote:
> > >
> > > >Having seen Giovanni Resta's numerical results, summing 7.5*10^11
> terms,
> > > >one may really ask if https://oeis.org/A318590 =
> > https://oeis.org/A222068
> > > >and if therefore A318590 should be merged into A222068.
> > > >
> > > >On Wed, Aug 29, 2018 at 9:26 PM Hugo Pfoertner <yae9911 at gmail.com>
> > wrote:
> > > >
> > > >> After summing 10^7 terms, I get 0.61685(016), which is a strange
> > > >> coincidence with https://oeis.org/A222068 (Pi/4)^2=0.616850275...
> > > >>
> > > >> On Wed, Aug 29, 2018 at 8:21 PM Hugo Pfoertner <yae9911 at gmail.com>
> > > wrote:
> > > >>
> > > >>> To continue with Ramanujan's questions: Question 770 dealing with
> > > >>> alternating sums of d(n)=A000005 is said to have been solved in two
> > > >>> articles in the Journal of the Indian Mathematical Society.
> Ramanujan
> > > >>> had asked to show that the infinite sum d(1) - d(3)/3+ d(5)/5 -
> > d(7)/7
> > > >>> + d(9)/9 - ... is a convergent series. Does somebody have access to
> > > the
> > > >>> articles or knows about a closed form solution? From a stupid
> > > summation
> > > >>> of 2*10^6 terms I get for the sum: 0.61684.., which is not in the
> > > OEIS,
> > > >>> so I created the draft https://oeis.org/draft/A318590
> > > >>>
> > > >>> Hugo Pfoertner
> > > >>>
> > > >>>
> > > >
> > > >--
> > > >Seqfan Mailing list - http://list.seqfan.eu/
> > > >
> > > >
> > >
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> > >
> >
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