The present problem is a continuation of the problem Riemann is back again . Now that you've explored some of the properties of the function where is the set of natural numbers such that converges, defined as
where is the set of all odd numbers, we would like to know how this function behaves as i.e., compute
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x ( n ) = ζ ( n ) 1 [ 1 n 1 + 3 n 1 + 5 n 1 + . . . ]
x ( n ) = ζ ( n ) 1 [ ζ ( n ) − ( 2 n 1 + 4 n 1 + 6 n 1 + . . . ) ]
x ( n ) = ζ ( n ) 1 [ ζ ( n ) − 2 n 1 ( 1 n 1 + 2 n 1 + 3 n 1 + . . . ) ]
x ( n ) = ζ ( n ) 1 [ ζ ( n ) − 2 n 1 ζ ( n ) ]
x ( n ) = 1 − 2 n 1
So:
n → ∞ lim x ( n ) = 1