Thursday, 21 August 2014

calculus - Computing limlimitsntoinftyBig(sumlimitsni=1sumlimitsnj=1frac1i2+j2fracpi2log(n)Big).

In the chatroom we discussed about the asymptotic of ni=1nj=11i2+j2, and if we think of the inverse tangent integral, it's easy to see that ni=1nj=11i2+j2Ti2(n)π2log(n) where I used
the well-known relation Ti2(x)Ti2(1/x)=π2sgn(x)log|x|. At this point, @robjohn posed the following limit



limn(ni=1nj=11i2+j2π2log(n))



that looks like a pretty tough limit. Using coth(z) one can see the limit is approximately π212,
but how about finding a way to precisely compute the limit?

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