sequences and series - What is $lim_{xto infty} 2sqrt{x}- sum_{n=1}^x {1over sqrt{n}}$?

I ask this because I noticed the partial sum $\sum_{n=1}^x {1\over \sqrt{n}}$ is very close to $2\sqrt{x}$, so close in fact that it appears their difference approaches a constant value, like $H_x$ and $\ln x$. However, when I put this limit as is into Wolfram, it said the limit diverged.

But, I found a way to transform the limit into an infinite sum, by using the transformation $f(x) = f(0) + \sum_{n=1}^x f(n) - f(n-1)$, an application of telescoping series to partial sums.

Thus,

$$\begin{align*}\lim_{x\to \infty} \left(2\sqrt{x}- \sum_{n=1}^x {1\over \sqrt{n}} \right) &= \lim_{x\to \infty} \left( 2\sqrt{0} + \sum_{n=1}^x \left( 2\sqrt{n} - 2\sqrt{n-1} \right) - \sum_{n=1}^x {1\over \sqrt{n}} \right) \\ &= \sum_{n=1}^{\infty} \left(2\sqrt{n} - 2\sqrt{n-1} -{1\over \sqrt{n}} \right) \end{align*}$$

This sum converges according to Wolfram by comparison test, and according to me by the integral test, but what does it converge to? It converges incredibly slowly; my best guess is $\approx 1.458$

Answer

May be, we could use generalized harmonic numbers since

$$\sum_{n=1}^x {1\over \sqrt{n}}=H_x^{\left(\frac{1}{2}\right)}$$ and use the asymptotics for large values of $n$ $$H_x^{\left(\frac{1}{2}\right)}=2 \sqrt{x}+\zeta \left(\frac{1}{2}\right)+\frac 1 {2\sqrt x}+O\left(\frac{1}{x^{3/2}}\right)$$