Let be the set of perfect powers. Evaluate the sum above to 2 decimal places. If you arrive at the conclusion that the sum diverges, enter your answer as 0.
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A lower bound is
S = ζ ( 2 ) + ζ ( 3 ) + ζ ( 5 ) + ζ ( 7 ) − ζ ( 6 ) − ζ ( 1 0 ) − ζ ( 1 4 ) − ζ ( 1 5 ) − ζ ( 2 1 ) − ζ ( 3 5 ) + ζ ( 3 0 ) + ζ ( 4 2 ) + ζ ( 7 0 ) + ζ ( 1 0 5 ) − ζ ( 2 1 0 ) − 1 ≈ 0 . 8 7 3 8 3 8 0 5
while an upper bound is
S ˉ = 1 − 3 × 4 1 − 7 × 8 1 − 8 × 9 1 − 1 5 × 1 6 1 − 2 4 × 2 5 1 − 2 6 × 2 7 1 − 3 2 × 3 2 1 − 3 5 × 3 6 1 − 4 8 × 4 9 1 − 6 3 × 6 4 1 − 8 0 × 8 1 1 − 9 9 × 1 0 0 1 ≈ 0 . 8 7 4 8 3 2 5 7
This is because
∑ k = 2 ∞ n k 1 = 1 − n 1 n 2 1 = n ( n − 1 ) 1 = n − 1 1 − n 1
∑ n = 2 ∞ [ n − 1 1 − n 1 ] = 1
while some of the terms are duplicates and can be removed.
Therefore, S ≈ 0 . 8 7 up to 2 decimal places.