# Inconsistency in summations using Wolfram|Alpha?

Posted 6 months ago
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 I have recently queried WolframAlpha this way... sum((1/(5*n+3)+1/(5*n+2))^2*(1/(5*n+4)+1/(5*n+1))^2*(3*n+1),-infinity<=n<=infinity) ...getting about -0.585507.But the result is very different with queries like these ones... sum((1/(5*n+3)+1/(5*n+2))^2*(1/(5*n+4)+1/(5*n+1))^2*(3*n+1),-1E20<=n<=1E20) sum((1/(5*n+3)+1/(5*n+2))^2*(1/(5*n+4)+1/(5*n+1))^2*(3*n+1),-1E20<=n<1E20) sum((1/(5*n+3)+1/(5*n+2))^2*(1/(5*n+4)+1/(5*n+1))^2*(3*n+1),-1E30<=n<=1E30) ... all giving about -1.09159.I am not a mathematician, I can't understand the result -0.585507. Answer
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Posted 6 months ago
 Is it possible that you are using the free version of WolframAlpha?If you scroll down just a bit from that answer do you see a pink warning "Standard computation time exceeded" and a button to click to try again using more computation time if you have a paid pro license?If you see all that then perhaps the explanation is that it ran out of time and used whatever partial calculation it had finished to provide an (incorrect) answer.Cannot be certain that this is exactly what happened behind the curtain.Using the full version of Mathematica, with fewer time limitations, this Sum[(1/(5*n+3)+1/(5*n+2))^2*(1/(5*n+4)+1/(5*n+1))^2*(3*n+1),{n,-Infinity,Infinity}] returns (35*Sqrt[5*(5 - 2*Sqrt)]*Pi - 45*Sqrt[5*(5 + 2*Sqrt)]*Pi + 100*Pi^2 + 16*Sqrt*Pi^2)/(25*(-5 + Sqrt)*(5 + Sqrt)) and then this N[%] approximates that complicated result as an approximate decimal value giving -1.09159 Answer
Posted 6 months ago
 Yes, I have web access to WolframAlpha free version only. Your explanation for the incorrect answer in my case is satisfying.(The real lesson for me is that also what appears in the first field of the answer-part of WolframAlpha can be affected by computation insufficient time.)Thank you for the support. Answer