Continued fraction expansion for coth(z)
We will show the proof for this continued fraction posted by @infseriesbot. e+1e−1=coth(12)=2+16+110+114+⋯ Update We add the proof of this result, which is bascially the same function evaluated at other point: 0=coth(−iπ2)=2−π26−π210−π214−⋯Recall the continued fraction representation for ratios of confluent hypergeometric functions of the type 0F1(c,z): Let (an) be a sequence of complex numbers defined by an=1(c+n−1)(c+n) where c is a complex constant such that c∉Z−∪{0}. Then 1+K∞n=1(anz1)=0F1(c,z)0F1(c+1,z)
Now, for this particular case, we will use the hypergemetric representation of sin(z) and cos(z) sin(z)=z0F1(32,−z24) sin(z)=0F1(12,−z24) Then tan(x)=sin(x)cos(x)=x−x33!+x55!−x77!+...1+x22!+x44!−x66!+...=x(1−x23!+x45!−x67!+...)1+x22!+x44!−x66!+...=x⋅0F1(32,−x24)0F1(12,−x24)=x1−x23−x25−x27−⋯ If we evaluate at x=iz2 tan(iz2)=iz22−−z26−−z210−−z214−⋯=iz2+z26+z210+z214−⋯⟹coth(z2)z=cot(iz2)iz=2+z26+z210+z214+⋯ If we evaluate at z=1 e+1e−1=coth(12)=2+16+110+114+⋯ Since the result is valid for al the complex plane, if we evaluate at z=−iπ then coth(−iπ2)=0 and 0=coth(−iπ2)=2−π26−π210−π214−⋯
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