Re: Mathematica doesn't recognize sum as e/(e+1)

On Jan 24, 3:28 am, Valeri Astanoff <astan...@xxxxxxxxx> writes:

VA> I wonder why such a powerful CAS as Mathematica 6
VA> doesn't recognize this sum as e/(e+1)
VA>
VA> In[1]:= Sum[-(2^(k+1)-1)*Zeta[-k]/k!, {k,0,Infinity}]
VA> Out[1]= Sum[((1 - 2^(1 + k))*Zeta[-k])/k!, {k,0,Infinity}]

True.

But consider, *none* of the Maple versions,
over decades (!) can calculate even THIS sum

sum((2^n+1)/n!, n=0..infinity);

sum((2^n+1)/n!,n = 0 .. infinity)

(!!)

Cheers,

Vladimir

--

Vladimir Bondarenko

VM and GEMM architect
Co-founder, CEO, Mathematical Director

http://www.cybertester.com/ Cyber Tester, LLC
http://maple.bug-list.org/ Maple Bugs Encyclopaedia
http://www.CAS-testing.org/ CAS Testing


On Jan 24, 3:28 am, Valeri Astanoff <astan...@xxxxxxxxx> wrote:
Good day,

I wonder why such a powerful CAS as Mathematica 6
doesn't recognize this sum as e/(e+1)

In[1]:= Sum[-(2^(k+1)-1)*Zeta[-k]/k!, {k, 0, Infinity}]
Out[1]= Sum[((1 - 2^(1 + k))*Zeta[-k])/k!, {k, 0, Infinity}]

In[2]:= %//N
During evaluation of In[2]:= NSum::nsumz: Some terms are zero.
        The algorithms are not very applicable. >>
Out[2]= 0.731059

In[3]:= E/(E+1)//N
Out[3]= 0.731059

v.a.

.

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