consider for n>=2
1/(n^3+1) < 1/n^3< 1/(n^3-1)
n^3+1 = (n+1)(n^2-n+1)
< (n+1)(n^2-n)
= (n+1)(n-1)n
1/(n^3+1) > 1/[(n-1)n(n+1)]
also
n^3-1 = (n-1)(n^2+n+1)
> (n-1)(n^2+n)
1/(n^3-1) < 1/[(n-1)n(n+1)]
1/[(i-1)i(i+1)] =(1/2){1/[(i-1)i] + 1/[i(i+1)] }
∑(i:2->n) 1/[(i-1)i(i+1)] = (1/2){1/2 + 1/[n(n+1)] }
lim(n->∞)∑(i:2->n) 1/[(i-1)i(i+1)] = 1/4
=> 1/3^3+1/2^3+....+1/n^3+... = 1+ 1/4 =5/4
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