Does $\lim\limits_{n→∞}\left(\sum\limits_{k=1}^n\frac1{\sqrt k}-2\sqrt n\right)$ exist?

Does the limit $$\displaystyle\lim_{n→∞}\left(\sum_{k=1}^n\dfrac1{\sqrt k}-2\sqrt n\right)$$ exist? With integral test I could make this bound: $$-2<\sum_{k=1}^n\frac1{\sqrt k}-2\sqrt n\le-1.$$ How would you tackle this problem?

Assuming you know the generalized harmonic numbers $$\sum_{k=1}^n \dfrac{1}{\sqrt{k}}=H_n^{\left(\frac{1}{2}\right)}$$ Now, using the asymptotics of them $$H_n^{\left(\frac{1}{2}\right)}=2 \sqrt{n}+\zeta \left(\frac{1}{2}\right)+\frac 1{2\sqrt{{n}}}+O\left(\frac{1}{n^{3/2}}\right)$$ making $$\sum_{k=1}^n \dfrac{1}{\sqrt{k}}-2\sqrt{n}=\zeta \left(\frac{1}{2}\right)+\frac 1{2\sqrt{{n}}}+O\left(\frac{1}{n^{3/2}}\right)$$ where $$\zeta \left(\frac{1}{2}\right)\approx -1.46035$$
Let us try for $$n=10$$; the "exact" calculation would give $$-1.30356$$ while the approximation would give $$-1.30224$$.
• So, the limit is $\zeta \left( \frac{1}{2} \right)$ ? – ARahman Oct 8 '18 at 10:45
Put $$a_n = \sum_{k=1}^n \frac1{\sqrt{k}} -2 \sqrt{n}.$$ We have $$a_{n+1} -a_n = \frac1{\sqrt{n+1}} -2(\sqrt{n+1} -\sqrt{n}) = \frac1{\sqrt{n+1}} -\frac{2}{\sqrt{n} + \sqrt{n+1}} = -\frac{\sqrt{n+1} -\sqrt{n}}{\sqrt{n+1}(\sqrt{n+1} +\sqrt{n})}< 0.$$ Then $$a_n$$ is decreasing sequence. Moreover $$a_n > -2$$. Hence there exists the limit $$\lim_{n\to \infty} a_n$$.