Note that if $f(x)=\frac{2x+\sqrt x}{2\pi}, x \ge 1$ then $f'(x)=\frac{1}{\pi}+\frac{1}{4\pi \sqrt x}$ so one has $\frac{1}{\pi} \le f'(x) \le \frac{5}{4\pi}$ and $f'$ decreasing so if $g(n)=f(n+1)-f(n), n \ge 1$, by the mean value theorem one has $n \le x_n \le n+1$ st $g(n)=f'(x_n)$ and of course this implies $\frac{1}{\pi} \le g(n) \le \frac{5}{4\pi}$ while also $g(n) \ge g(n+1)$ since $f'$ decreasing
Now with $e(x)=e^{2\pi ix}$ and $S_n=\sum_{k=1}^ne(f(k))$ we have that $\sum_{k=1}^n \cos (2k+\sqrt k)=\Re S_n$ so if we prove that $|S_n| \le C$ (hence $|\Re S_n| \le C$) we can apply summation by parts to conclude that the series $\sum_{n=1}^{\infty} \frac{\cos (2n+\sqrt{n})}{\sqrt{n}}$ converges
Noting that $\frac{1}{1-e(x)}=\frac{ie^{-i\pi x}}{2\sin \pi x}=\frac{1}{2}(1+i\cot \pi x)$, and
$e(f(k))(1-e(g(k)))=e(f(k))-e(f(k+1))$,
we have $e(f(k))=(e(f(k))-e(f(k+1)))(\frac{1}{2}(1+i\cot \pi g(k))$, so
$$S_n=\sum_{k=1}^ne(f(k))=\sum_{k=1}^{n-1}(e(f(k))-e(f(k+1)))(\frac{1}{2}(1+i\cot \pi g(k))+e(f(n))=$$
$$=e(f(1))(\frac{1}{2}(1+i\cot \pi g(1))+i/2\sum_{k=2}^{n-1}e(f(k))(\cot \pi g(k)-\cot \pi g(k-1))+e(f(n))(1/2-i\cot \pi g(n-1))$$
(last equality obtained by grouping on $e(f(k))$)
So if we take absolute values we get (noting that $1 \le \pi g(k) \le 5/4 <\pi/2$, so the cotangent is positive and decreasing):
$$|S_n| \le 1+\frac{\cot \pi g(1)+ \cot \pi g(n-1)}{2}+\frac{1}{2}\sum_{k=2}^{n-1}|(\cot \pi g(k)-\cot \pi g(k-1))| =$$
$$= 1+\frac{\cot \pi g(1)+ \cot \pi g(n-1)}{2}+\frac{1}{2}(\cot \pi g(n-1)-\cot \pi g(1))=1+\cot \pi g(n-1) \le 1+\cot 1=C$$
since the sum telescopes as $g_n$ decreasing so we can remove the absolute values and we are done as the partial sums $\sum_{k=1}^n \cos (2k+\sqrt k)$ are uniformly bounded so the series converges
Note that the same proof with small modifications shows that $\sum_{k=1}^n\cos (ak+bk^c)$ is uniformly bounded for $a \ne 2m\pi, c<1$ (with a bound depending on $a,b,c$ of course) as taking $f(x)=\frac{ax+bx^c}{2\pi}, x \ge 1$ we have $f'$ monotonic and $f' \to \frac{a}{2\pi}, x \to \infty$ hence if $\frac{a}{2\pi}$ is not an integer, we get that $||f'(x)|| > d>0$ for $x >N(a,b,c)$ large enough where $||.||$ is the distance to the nearest integer so the cotangent technique applies for the sum from $N(a,b,c)$ on while the rest of the terms are finite (if $b=0$ the result is direct from the geometric series summation).