∞∑n=1csc2(nπ5)n3=∞∑m=010∑n=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=02,3,4,6,7,8,9∑n=11(10m+n)3csc2(nπ5)
We have eventually,
π225[2,4,5∑i=1csc2((2i−1)π5)∞∑m=11(10m+(2i−1))3+184∑i=1csc2(2iπ5)∞∑m=11(5m+i)3+2,3,4,6,7,8,9∑n=11n3csc2(nπ5)]
This will converge and can be solved, numerically.