Re: Infinite series
- To: mathgroup at smc.vnet.net
- Subject: [mg105701] Re: [mg105683] Infinite series
- From: "Tony Harker" <a.harker at ucl.ac.uk>
- Date: Wed, 16 Dec 2009 06:16:55 -0500 (EST)
- References: <200912151227.HAA15202@smc.vnet.net>
FullSimplify will do it, or you can sneak round the problem with
Sum[Apart[1/((2*m - 3)^2*(2*m - 1)*(2*m + 1)^2)], {m, 2, Infinity,
2}] - Sum[
Apart[1/((2*m - 3)^2*(2*m - 1)*(2*m + 1)^2)], {m, 1, Infinity,
2}] // Simplify
Tony Harker
Dr A.H. Harker
Department of Physics and Astronomy
University College London
Gower Street
London
WC1E 6BT
Tel: (44)(0) 2076793404
E: a.harker at ucl.ac.uk
EDUCATION, n. That which discloses to the wise and disguises from the
foolish their lack of understanding. (Ambrose Bierce, The Devil's
Dictionary, 1911)
]-> -----Original Message-----
]-> From: Dr. C. S. Jog [mailto:jogc at mecheng.iisc.ernet.in]
]-> Sent: 15 December 2009 12:27
]-> To: mathgroup at smc.vnet.net
]-> Subject: [mg105683] Infinite series
]->
]-> Hi:
]->
]-> We have the following identity:
]->
]-> \sum_{m=1}^{infinity} (-1)^m/((2m-3)^2*(2m-1)*(2m+1)^2)=-Pi/32.
]->
]-> When we type the command,
]->
]-> In[1]:=Sum[(-1)^m/((2*m-3)^2*(2*m-1)*(2*m+1)^2),{m,Infinity}]
]->
]-> we get
]-> 2 1 1
]-> -16 Pi + 2 Pi - HurwitzZeta[2, -(-)] - Zeta[2, -]
]-> 4 4
]-> Out[1]= --------------------------------------------------
]-> 512
]->
]->
]-> The command Simplify[%] does not simplify it further.
]->
]-> I am sure the above expression must be equal to -Pi/32, but
]-> a user would prefer this answer than the above one.
]->
]-> Thanks and regards
]->
]-> C. S. Jog
]->
]->
]->
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- References:
- Infinite series
- From: "Dr. C. S. Jog" <jogc@mecheng.iisc.ernet.in>
- Infinite series