Evaluate: $\lim_{n\to\infty}\left({2\sqrt n}-\sum_{k=1}^n\frac1{\sqrt k}\right)$ How to find $\lim\limits_{n\to\infty}\left({2\sqrt n}-\sum\limits_{k=1}^n\frac1{\sqrt k}\right)$ ?
And generally does the limit of the integral of $f(x)$ minus the sum of $f(x)$ exist?
How to prove that and find the limit?

Use $\sqrt{n} = \sum_{k=1}^n \left( \sqrt{k} - \sqrt{k-1} \right)$, then
$$
\begin{eqnarray}
  2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} &=& \sum_{k=1}^n \left( 2 \sqrt{k} - 2 \sqrt{k-1} - \frac{1}{\sqrt{k}} \right) = \sum_{k=1}^n \frac{1}{\sqrt{k}}  \left( \sqrt{k}-\sqrt{k-1} \right)^2\\
 &=& \sum_{k=1}^n \frac{1}{\sqrt{k}} \left( \frac{(\sqrt{k}-\sqrt{k-1})(\sqrt{k}+\sqrt{k-1})}{(\sqrt{k}+\sqrt{k-1})} \right)^2 \\
  &=& \sum_{k=1}^n \frac{1}{\sqrt{k} \left(\sqrt{k}+\sqrt{k-1}\right)^2} 
\end{eqnarray}
$$
This shows the limit does exist and $\lim_{n \to \infty} \left( 2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} \right) = \sum_{k=1}^\infty  \frac{1}{\sqrt{k} \left(\sqrt{k}+\sqrt{k-1}\right)^2}$.
The value of this sums equals $-\zeta\left(\frac{1}{2} \right) \approx 1.4603545$. This value is found by other means, though:
$$
  2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} = 2 \sqrt{n} - \left( \zeta\left(\frac{1}{2}\right) - \zeta\left(\frac{1}{2}, n+1\right)\right) \sim -\zeta\left(\frac{1}{2}\right) - \frac{1}{2\sqrt{n}} + o\left( \frac{1}{n} \right) 
$$ 
