Calculate below limit
lim
Answer
As a consequence of Euler's Summation Formula, for s > 0, s \neq 1 we have
\sum_{j =1}^n \frac{1}{j^s} = \frac{n^{1-s}}{1-s} + \zeta(s) + O(|n^{-s}|),
where \zeta is the Riemann zeta function.
In your situation, s=1/2, so
\sum_{j =1}^n \frac{1}{\sqrt{j}} = 2\sqrt{n} + \zeta(1/2) + O(n^{-1/2}) ,
and we have the limit
\lim_{n\to \infty} \left( \sum_{j =1}^n \frac{1}{\sqrt{j}} - 2\sqrt{n} \right) = \lim_{n\to \infty} \big( \zeta(1/2) + O(n^{-1/2}) \big) = \zeta(1/2).
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