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?
 A: If (as in your case) the function $f$ decreases monotonically to zero, then the limit of (integral minus sum) exists and is at most $f(1)$. You can see that by drawing a picture: draw the graph of $f$, and little rectangles jutting up over it, so the area under the graph is the integral, and the area of the rectangles gives the sum. If you push all the little excesses (the bits of rectangle above the curve) over to the left, they don't fill up the first rectangle.  
A: A rather brute force approach, but by the Euler-Maclaurin formula,
$$2\sqrt n-\sum_{k=1}^n\frac1{\sqrt k}=-\zeta(1/2)-\frac1{2\sqrt n}+R_n$$
where
$$|R_n|\le\frac1{8n^{3/2}}$$
And so, as $n\to\infty$, we get
$$\lim_{n\to\infty}\left(2\sqrt n-\sum_{k=1}^n\frac1{\sqrt k}\right)=-\zeta(1/2)$$
A: 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:
$$
  2 \sqrt{n} - \sum_{k=1}^n \frac{1}{\sqrt{k}} = 2 \sqrt{n} - \left( \zeta\left(\frac{1}{2}\right) - \zeta\left(\frac{1}{2}, n+1\right)\right) \sim -\zeta\left(\frac{1}{2}\right) - \frac{1}{2\sqrt{n}} + o\left( \frac{1}{n} \right) 
$$ 
