TITLE: Convergence of $ \sum_{n=1}^{\infty} \frac{\cos (2n+\sqrt{n})}{\sqrt{n}} $
QUESTION [7 upvotes]: 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.

REPLY [2 votes]: (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.