Answer
Answer : $\frac{1}{2}$
Work Step by Step
This is basically telescoping series after we expand a few terms such that
$\sum_{n = 4}^{k} (\frac{1}{\sqrt{n}} = \frac{1}{\sqrt{n+1}})=(\frac{1}{\sqrt{4}}-\frac{1}{\sqrt{5}}) + (\frac{1}{\sqrt{5}} - \frac{1}{\sqrt{6}}) + (\frac{1}{\sqrt{6}}-\frac{1}{\sqrt{7}}) + \cdots+(\frac{1}{\sqrt{k-1}} - \frac{1}{\sqrt{k}}) + (\frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}})$
We can see that the terms in between cancel out and the series left with
$\sum_{n=4}^{k} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}}) = \frac{1}{\sqrt{4}} - \frac{1}{\sqrt{k + 1}}$
Now by limiting $k$ approaches $\infty$, we finally get that
$\lim_{k\to\infty}\sum_{n=4}^{k} (\frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}}) = \frac{1}{\sqrt{4}} - 0 = \frac{1}{2}$
