Calculate below limit
limn→∞(n∑i=11√i−2√n)
Answer
As a consequence of Euler's Summation Formula, for s>0, s≠1 we have
n∑j=11js=n1−s1−s+ζ(s)+O(|n−s|),
where ζ is the Riemann zeta function.
In your situation, s=1/2, so
n∑j=11√j=2√n+ζ(1/2)+O(n−1/2),
and we have the limit
limn→∞(n∑j=11√j−2√n)=limn→∞(ζ(1/2)+O(n−1/2))=ζ(1/2).
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