Evaluate $a ;b$ values which satisfy this equality : $$b = \lim_{n \to + \infty } n^a \sum_{k = 1}^n {\frac{1}{{\sqrt k }}}$$
My solution
$$ \sqrt n \le \sum\limits_{k = 1}^n {\frac{1}{{\sqrt k }}} \le 2\sqrt n \Rightarrow n^{\frac{1}{2} + a} \le n^a \sum\limits_{k = 1}^n {\frac{1}{{\sqrt k }} \le 2} n^{\frac{1}{2} + a} $$ $$ b = \mathop {\lim }\limits_{n \to + \infty } n^{\frac{1}{2} + a} \le \mathop {\lim }\limits_{n \to + \infty } n^a \sum\limits_{k = 1}^n {\frac{1}{{\sqrt k }} \le \mathop {\lim }\limits_{n \to + \infty } 2} n^{\frac{1}{2} + a} = b $$
$$ \Rightarrow \mathop {\lim }\limits_{n \to + \infty } n^{\frac{1}{2} + a} = 0 $$ $$ \Rightarrow \frac{1}{2} + a \le - 1 \Leftrightarrow a \le - \frac{3}{2} $$
Is this solution correct?
Let $S_n = \displaystyle\sum_{k=1}^n \dfrac1{\sqrt{k}}$. We then have $S_n = \displaystyle \int_{1^-}^{n^+} \dfrac{dA(x)}{\sqrt{x}}$ where $A(x) = \sum_{k \leq x} 1 = \lfloor x \rfloor$. Integrating by parts, we obtain $$S_n = \dfrac{A(n)}{\sqrt{n}} + \dfrac12 \int_{1^-}^{n^+} \dfrac{A(x)}{x^{3/2}}dx = \sqrt{n} + \dfrac12 \int_{1^-}^{n^+}\dfrac{x-\{x\}}{x^{3/2}}dx=2\sqrt{n}-1 - \underbrace{\dfrac12\int_{1^-}^{n^+}\dfrac{\{x\}}{x^{3/2}}dx}_{\mathcal{O}(1) \text{ since }\{x\} \in [0,1)}$$ Hence, $S_n = 2\sqrt{n} + \mathcal{O}(1)$. Hence, for $\lim_{n \to \infty} n^a S_n$ to exists, we would need $a \leq -1/2$. For the limit to be a non-zero number, we would need $a=-1/2$.