Sum of a series: Difference between revisions
→{{header|Lucid}}
Line 138:
sum(1,1000, function(x) { return 1/(x*x) } ) // 1.64393456668156
=={{header|Lucid}}==
series = ssum asa n >= 1000
where
num = 1 fby num + 1;
ssum = ssum + 1/(num * num)
end;
=={{header|Perl}}==
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