From: supergems on 5 Oct 2009 04:31 Hi all, is there a way to calculate this series: '\GS(N=2,\oo,2^(N+1)/(N-1)!)' with HP 50g? The answer is '4*EXP(2)-4' . Regards, Simone. From: Virgil on 5 Oct 2009 05:20 In article <5a49ea19-60d9-46fe-8887-095e056c4a97(a)j28g2000vbl.googlegroups.com>, supergems wrote: > Hi all, is there a way to calculate this series: > > '\GS(N=2,\oo,2^(N+1)/(N-1)!)' > > with HP 50g? > The answer is '4*EXP(2)-4' . > > Regards, > Simone. One can, with a little imagination, express your series in terms of the series '\GS(N=0,\oo,2^(N)/N!)', which calculus tells us evaluates to 'EXP(2)' From: sc on 13 Oct 2009 22:44 On Oct 5, 4:31 am, supergems wrote:> Hi all, is there a way to calculate this series: > > '\GS(N=2,\oo,2^(N+1)/(N-1)!)' > > with HP 50g? > The answer is '4*EXP(2)-4' . > > Regards, > Simone. Rewrite: Sum[(2^(i+1))/(i!),{i,2,Infinity}] is equivalent to 4*Sum[(2^i)/(i!),{i,1,Infinity}] (change the index, factor out the 2^2 from the numerator) = 4*(e^2-1). [Taylor series of e^x, except it's missing the initial term of 1] I couldn't figure out how to do it on the 50g. Humans are smarter than calculators anyway. S.C.