## How to compute sum(log(1-1/n^2),n,2,oo)?

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 How to compute sum(log(1-1/n^2),n,2,oo)? Alberto Verga 4/16/14 3:19 AM `n = var('n')sum(log(1-1/n^2),n,2,oo)`do not give the answer -log(2), but a 0*infinity error. Re: [sage-support] How to compute sum(log(1-1/n^2),n,2,oo)? jori.ma...@uta.fi 4/16/14 3:29 AM limit(sum(log(1-1/n^2),n,2,x), x=oo) -- Jori Mäntysalo Re: [sage-support] How to compute sum(log(1-1/n^2),n,2,oo)? Alberto Verga 4/16/14 5:58 AM OK, but why we need to use "limit" to compute a convergent sum...?For instance    sum(1/n^2,n,2,oo)gives pi^2/6-1but     limit(sum(1/n^2,n,2,x), x=oo)do not give a result... Re: [sage-support] How to compute sum(log(1-1/n^2),n,2,oo)? kcrisman 4/16/14 10:33 AM On Wednesday, April 16, 2014 8:58:34 AM UTC-4, Alberto Verga wrote:OK, but why we need to use "limit" to compute a convergent sum...?For instance    sum(1/n^2,n,2,oo)gives pi^2/6-1but     limit(sum(1/n^2,n,2,x), x=oo)do not give a result...This is interesting.  In Sage's Maxima:(%i1) load(simplify_sum);(%i9) simplify_sum(sum(1/x^2,x,2,inf));                                      2                                   %pi(%o9)                              ---- - 1                                    6(%i10) sum(log(1-1/x^2),x,2,inf);                               inf                               ====                               \             1(%o10)                          >    log(1 - --)                               /              2                               ====          x                               x = 2(%i11) simplify_sum(sum(log(1-1/x^2),x,2,inf));                                     inf + 2(%o11)                         log(-----------)                                   2 (inf + 1)Which I agree is not a great answer.  Using simpsum does the same thing.  I'm not sure if this is expected behavior in Maxima, though? Re: How to compute sum(log(1-1/n^2),n,2,oo)? Robert Dodier 4/16/14 12:34 PM On 2014-04-16, kcrisman wrote: > (%i11) simplify_sum(sum(log(1-1/x^2),x,2,inf)); >                                      inf + 2 > (%o11)                         log(-----------) >                                    2 (inf + 1) > > Which I agree is not a great answer.  Using simpsum does the same thing. >  I'm not sure if this is expected behavior in Maxima, though? Well, %o11 is certainly incorrect, although it is perhaps not too surprising. simplify_sum makes use of several methods, including Gosper's and Zeilberger's methods, which (if I recall correctly) are applicable to indefinite finite sums (i.e., upper limit is a symbol, say m, such that m < inf). My guess is that such a method was applied mistakenly to %i11. Of course, that's a bug in simplify_sum. For the record: (%i19) simplify_sum (sum (log (1 - 1/x^2), x, 2, m)); (%o19) log((m+1)/(2*m)) (%i20) limit (%, m, inf); (%o20) -log(2) best, Robert Dodier Re: How to compute sum(log(1-1/n^2),n,2,oo)? kcrisman 4/16/14 5:50 PM Thanks, Robert.  Yes, the latter is essentially why Jori's answer works, I believe. - kcrisman Re: How to compute sum(log(1-1/n^2),n,2,oo)? kcrisman 4/17/14 11:17 AM And Maxima upstream already has committed a fix, apparently - great news.  Thanks for the report!