In the chatroom we discussed about the asymptotic of n∑i=1n∑j=11i2+j2, and if we think of the inverse tangent integral, it's easy to see that n∑i=1n∑j=11i2+j2≈Ti2(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→∞(n∑i=1n∑j=11i2+j2−π2log(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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