Convergence of $ \sum_{n=1}^{\infty} \frac{\cos (2n+\sqrt{n})}{\sqrt{n}} $ As you undestand from the topic title, I am wondering how to determine whether the series
$$ \sum_{n=1}^{\infty} \frac{\cos (2n+\sqrt{n})}{\sqrt{n}}  $$
converges or not.
Previously there have been many similar problems and I tried them all did not work.
[Link1]:(Convergence of $\sum \frac{ \cos\sqrt n} {\sqrt n}$) Estimating the number of n to identify a Cauchy lower bound is a good approach, but  when inside the $\cos$ is $2n+\sqrt{n}$, the range of $n$ is not as much as expected failed to estimate the lower boundary.
[Link2]:(Does $\sum_{n=1}^\infty \frac{\cos{(\sqrt{n})}}{n}$ converge?)I followed the idea of Robert Z's answer. Although its corresponding integral is convergent, the summation does not converge due to the 1st order term in the first step of estimating $|f(n)-f(x)|$.
Any help may yield help and look forward to your reply.
 A: 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).
A: Let $f(x)=\frac{e^{i\sqrt{x}}}{\sqrt{x}}$. The point is that $f\to 0,f'\to 0$ and $f''=O(x^{-3/2})$.
Doing two partial summations we get that $$\sum_{n=1}^N e^{2in}f(n)= f(N)\sum_{n=1}^N e^{2in}+\sum_{n=1}^{N-1} (\sum_{m=1}^n e^{2im}) (f(n)-f(n+1)) $$
$$ = o(1)+\sum_{n=1}^{N-1} \frac{e^{2in}-1}{1-e^{-2i}} (f(n)-f(n+1))$$
$$ = o(1)+C+\sum_{n=1}^{N-1} \frac{e^{2in}}{1-e^{-2i}} (f(n)-f(n+1))$$
$$ = o(1)+C+\underbrace{(f(N-1)-f(N))\sum_{n=1}^{N-1} \frac{e^{2in}}{1-e^{-2i}}}_{o(1)}\\\qquad+ \sum_{n=1}^{N-2} \underbrace{\frac{\sum_{m=1}^n e^{2in}}{1-e^{-2i}}}_{O(1)}\underbrace{ (f(n)-2f(n+1)+f(n+2))}_{O(n^{-3/2})}$$
converges as $N\to \infty$.

Whence $\sum_{n=1}^\infty \Re(e^{2in}f(n))$ converges.

A: (Disclaimer: This is essentially a paraphrasing of @reuns's answer.)
For $x > 0$, define
\begin{align*}
f_1(x)
&= \left( \int_{[1,x]} e^{2it} \, \mathrm{d}\lfloor t \rfloor + \frac{e^{2i}}{e^{2i} - 1} \right)
= \frac{e^{2i}}{e^{2i} - 1} \, e^{2i\lfloor x\rfloor}, \\
f_2(x)
&= \left( \int_{1}^{x} f_1(t) \, \mathrm{d}t \right)
= \frac{e^{2i}}{e^{2i} - 1} \left( \frac{e^{2i\lfloor x\rfloor} - e^{2i}}{e^{2i}-1} + (x - \lfloor x \rfloor) e^{2i\lfloor x \rfloor} \right), \\
g(x) &= \frac{e^{i\sqrt{x}}}{\sqrt{x}}.
\end{align*}
Then both $f_1$ and $f_2$ are bounded. Moreover, it is easy to verify that
$$ g(x) = \mathcal{O}\bigl(x^{-1/2}\bigr), \qquad 
g'(x) = \mathcal{O}\bigl(x^{-1}\bigr), \qquad
g''(x) = \mathcal{O}\bigl(x^{-3/2}\bigr) \qquad \text{as } x \to \infty. $$
On the other hand, using integration by parts twice,
\begin{align*}
\sum_{n=1}^{N} \frac{e^{2in + i\sqrt{n}}}{\sqrt{n}}
&= \int_{[1, N]} e^{2it}g(t) \, \mathrm{d}\lfloor t \rfloor \\
&= \left[ f_1(t) g(t) \right]_{1^-}^{N} - \left[ f_2(t) g'(t) \right]_{1}^{N} + \int_{1}^{N} f_2(t) g''(t) \, \mathrm{d}t\
\end{align*}
The last expression converges as $N\to\infty$, and so, the desired conclusion follows.
