((1/(6n+1)^21/(6n+2)^2+1/(6n+4)^21/(6n+5)^2)/((1/(6n+1)^2+1/(6n+2)^21/(6n+4)^21/(6n+5)^2))) is worth 1/3 in the limit, according to wolframalpha.
On reflection, possibly a bit unfair to wolframalpha. The exact expression is:
(sum n=0 to infinity ((1/(6n+1)^21/(6n+2)^2+1/(6n+4)^21/(6n+5)^2))/(sum n=0 to infinity (1/(6n+1)^2+1/(6n+2)^21/(6n+4)^21/(6n+5)^2))), [for the precise wolframalpha syntax required see comment 4]
so the original version was open to misinterpretation.
Edited on October 13, 2012, 10:24 am

Posted by broll
on 20121012 10:31:35 