Is it possible to calculate the following infinite sum in a closed form? If yes, please point me to the right direction.
∞∑n=1(n arccot n−1)
Answer
ncot−1(n)−1=ntan−1(1n)−1=n∫1/n0dx1+x2−1=−n∫1/n0x2dx1+x2=−∫10x2dxn2+x2
Using formula (9) from this answer and substituting z↦ix, we get
∞∑n=11n2+x2=πcoth(πx)2x−12x2
Combining (1) and (2) yields
∞∑n=1(ncot−1(n)−1)=12∫10(1−πxcoth(πx))dx=12∫10(1−πx(1+2e−2πx1−e−2πx))dx=2−π4−π∫10xe−2πx1−e−2πxdx=2−π4−π∫10x(∞∑n=1e−2πnx)dx=2−π4−π∞∑n=1(1(2πn)2−(12πn+1(2πn)2)e−2πn)=2−π4−π24−12log(1−e−2π)+14πLi2(e−2π)=12+17π24−12log(e2π−1)+14πLi2(e−2π)
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