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Thursday, December 29, 2022

Taking Out The Explosives First

 n=1csc2(nπ5)n3=m=010n=11(10m+n)3csc2(nπ5)

As the sum at n=5 and n=10 are positive and negative infinite sum, they cancel and we are left with,

  n=1csc2(nπ5)n3=m=02,3,4,6,7,8,9n=11(10m+n)3csc2(nπ5)

We have eventually,

π225[2,4,5i=1csc2((2i1)π5)m=11(10m+(2i1))3+184i=1csc2(2iπ5)m=11(5m+i)3+2,3,4,6,7,8,9n=11n3csc2(nπ5)]

This will converge and can be solved, numerically.