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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 <simone.cerica(a)gmail.com> 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 <simone.cer...(a)gmail.com> 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.
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