First of all : $$ \prod_{p=1}^{n}{\left(1+\frac{1}{4p}\right)}\prod_{p=0}^{n}{\left(1-\frac{1}{4p+3}\right)}=\frac{4\Gamma\left(\frac{7}{4}\right)\Gamma\left(n+\frac{5}{4}\right)\Gamma\left(n+\frac{3}{2}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)\Gamma\left(n+1\right)\Gamma\left(n+\frac{7}{4}\right)}\underset{n\to +\infty}{\longrightarrow}\frac{4\Gamma\left(\frac{7}{4}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)} $$
Let $ n $ be a positive integer, we have the following : \begin{aligned}\sum_{p=1}^{2n+1}{\left(-1\right)^{p}\ln{\left(1+\frac{1}{2p}\right)}}&=\sum_{k=1}^{n}{\ln{\left(1+\frac{1}{4p}\right)}}-\sum_{k=0}^{n}{\ln{\left(1+\frac{1}{4p+2}\right)}}\\&=\ln{\left(\prod_{k=1}^{n}{\left(1+\frac{1}{4p}\right)}\prod_{k=0}^{n}{\left(1-\frac{1}{4p+3}\right)}\right)} \end{aligned}
Thus, $ \sum\limits_{p\geq 0}{\left(-1\right)^{p}\ln{\left(1+\frac{1}{2p}\right)}} $ converges, and its sum values $ \ln{\left(\frac{4\Gamma\left(\frac{7}{4}\right)}{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)}\right)}\cdot $
Now, since $ \sum\limits_{n\geq 1}{\frac{\left(-1\right)^{n-1}}{\left(2n+1\right)^{s}}} $ converges absolutly, for any $ s>1 $, we can prove that the family $ \left(\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}\right)_{\left(p,n\right)\in\mathbb{N}^{*}\times\mathbb{N}^{*}\setminus\left\lbrace 1\right\rbrace} $ is summable, and hence, thanks to Fubini's theorem, we can write the following : \begin{aligned}\sum_{n=2}^{+\infty}{\frac{1-\beta\left(n\right)}{n}}=\sum_{n=2}^{+\infty}{\sum_{p=1}^{+\infty}{\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}}}=\sum_{p=1}^{+\infty}{\sum_{n=2}^{+\infty}{\frac{\left(-1\right)^{p-1}}{n\left(2p+1\right)^{n}}}}\end{aligned}
After switching the summations, adding the first term of the sum, recognising logarithm's series expansion, we get that : $$ \sum_{n=1}^{+\infty}{\frac{1-\beta\left(n\right)}{n}}=\sum_{p=1}^{+\infty}{\left(-1\right)^{p-1}\ln{\left(1+\frac{1}{2p}\right)}}=\ln{\left(\frac{3\sqrt{\pi}\Gamma\left(\frac{5}{4}\right)}{4\Gamma\left(\frac{7}{4}\right)}\right)} $$
Best Answer
Suppose we are given a small $\epsilon\gt 0$. We show that we can choose $n$ such that $\sum_1^n\sqrt{k}$ is within $\epsilon$ of an integer. Relatively simple estimates are used.
For take $N\gt 1/\epsilon$. Then the numbers $$\sqrt{N^4+1},\quad \sqrt{N^4+2},\quad\text{and so on up to}\quad \sqrt{N^4+2N} $$ are greater than $N^2$ but within $\epsilon$ of $N^2$. Thus each of them is "nearly" an integer. To see this, note that $\left(N^2+\frac{1}{N}\right)^2 \gt N^4 +2N$.
Moreover, these numbers have fractional parts that add up to more than $1$. This is fairly straightforward, since the smallest fractional part is approximately $\frac{1}{2N}$.
So however far $\sum_1^{N^4} \sqrt{k}\,\,$ may be from an integer, one of the sums to $n=N^4+i$, where $1\le i\le 2N$, must come within $\epsilon$ of an integer.
Remark: It may be interesting to ask how much better one can do than $M\approx \frac{1}{\epsilon^4}$ to be sure that there is an $n\le M$ such that $\sum_1^n\sqrt{k}$ is within $\epsilon$ of an integer. Presumably much better! That is where more sophisticated ideas such as Euler-Maclaurin may be useful.