Does $\sum_{n=1}^\infty \frac{\cos{(\sqrt{n})}}{n}$ converge? The series is:
$$\sum_{n=1}^\infty \frac{\cos(\sqrt{n})}{n}$$ 
Considering it isn't always positive, I replace $\frac{\cos{\sqrt{n}}}{n}$ with its absolute value and I find that:
$$\vert \frac{\cos{\sqrt{n}}}{n}\vert\gt \frac{\cos^2{\sqrt{n}}}{n}=\frac{\ 1+\cos{2\sqrt{n}}}{2n}=\frac{1}{2n}+\frac{\cos{2\sqrt{n}}}{2n}$$
if $\sum_{n=1}^\infty \vert\frac{\cos{\sqrt{n}}}{n}\vert $   converges, then 
$\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ converges.Using Comparison test,we can draw the conclusion that  $\sum_{n=1}^\infty\frac{1}{2n}$  converges , which is impossible.
So I get that $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$  absolutely diverges. But I can't figure out whether  $\sum_{n=1}^\infty \frac{\cos{\sqrt{n}}}{n}$ converges or not.
I have tried Dirichlet's test, but I can't figure out whether $$S_{n}=\sum_{k=1}^n \cos{\sqrt{k}}$$ is bounded.
(This is my first time to ask question.Maybe there exist some mistakes in my conclusion.Thanks. :) 
 A: Notations: $\lfloor x \rfloor$ is the floor function, $\{x\}$ is the fractional part of $x$ so that $x=\lfloor x\rfloor + \{x\}$.
Applying partial summation with $f(x)=\frac{\cos(\sqrt x)}x$,
$$\begin{align}
\sum_{n=1}^N \frac{\cos(\sqrt n)}n&=\int_{1-}^N f(x)d\lfloor x \rfloor \\
&=f(x)\lfloor x\rfloor \Big\vert_{1-}^N-\int_{1-}^N f'(x)\lfloor x\rfloor dx\\
&=f(N)(N-\{N\})-\int_1^N xf'(x)dx+\int_1^N\{x\}f'(x)dx\\
&=Nf(N)-f(1)-\int_1^Nxf'(x)dx+\int_1^N\{x\}f'(x)dx+f(1)-\{N\}f(N).
\end{align}
$$
From integration by parts, the sum of first three terms is
$$
\int_1^{N}f(x)dx
$$
Thus, we have the following as $N\rightarrow\infty$,
$$
\sum_{n=1}^{\infty}\frac{\cos(\sqrt n)}n=\int_1^{\infty}f(x)dx+\int_1^{\infty}\{x\}f'(x)dx+\cos 1.
$$
It is easy to see that the integrals converge.
A: This development is similar to that of user i707107 but it is more detailed and avoids the strange expression $d \lfloor x \rfloor$. 
Letting
$$c_{k} = c(k) = \frac{\cos(\sqrt{k})}{k}$$
we attempt to find an integral representation for the partial sum 
$$s_n = \sum_{k=1}^n c_{k}$$
The formulas for partial summation are
$$\sum_{k=1}^n a_{k} b_{k} = A_{n} b_{n} + \sum_{k=1}^{n-1} A_{k}(b_{k}-b_{k+1})$$
$$A_{k} = \sum_{i=1}^k a_{i}$$
Letting $a_{k}=1, b_{k} = c_{k}$ we have $A_{k} = k$.
Now comes the trick which introduces an integral: we have
$$(b_{k}-b_{k+1}) = - \int_{k}^{k+1} c'(x)\;dx$$
and, what's more, the factor $k$ can be incorporated in the integral:
$$k (b_{k}-b_{k+1}) = - k \int_{k}^{k+1} c'(x)\;dx = - \int_{k}^{k+1} \lfloor x\rfloor c'(x)\;dx $$
Where $\lfloor x\rfloor$ is the floor function.
Hence, using 
$$\lfloor x\rfloor = x -\{x\} $$
where $\{x\}$ is the fractional part of $x$, the partial sum becomes
$$s_{n} = n c_n -\sum_{k=1}^{n-1} \int_{k}^{k+1} \lfloor x\rfloor c'(x)\;dx \\
=  n c_n -\int_{1}^{n} x c'(x)\;dx + \int_{1}^{n} \{x\} c'(x)\;dx$$ 
By partial integration of the first integral the term $n c_{n}$ drops out and we get
$$s_{n} = \cos(1) +\int_{1}^{n} c(x)\;dx + \int_{1}^{n} \{x\} c'(x)\;dx$$ 
The first integral can be solved explicitly
$$\int_{1}^{n} c(x)\;dx = 2 \text{Ci}\left(\sqrt{n}\right)-2 \text{Ci}(1)$$
Where $\text{Ci}$ is the integral cosine.
The second integral is absolutely convergent and can be very roughly estimated thus (notice that $0\le \{x\} \lt 1$)
$$|i_2| =  |\int_{1}^{n} \{x\} c'(x)\,dx| <=  \int_{1}^{n}| \{x\} c'(x)|\,dx \\
<=  \int_{1}^{n} |\{x\}| |c'(x)|\,dx 
<=\int_{1}^{n} | c'(x) | \,dx$$
Now 
$$| c'(x) | = |\frac{\cos \left(\sqrt{x}\right)}{x^2}+\frac{\sin \left(\sqrt{x}\right)}{2 x^{3/2}}| \\
\leq | \frac{\sin \left(\sqrt{x}\right)}{2 x^{3/2}}| +|\frac{\cos \left(\sqrt{x}\right)}{x^2}| \leq \frac{1}{2 x^{3/2}}+\frac{1}{x^2}$$
Hence
$$| i_2| <=\int_1^n \left(\frac{1}{2 x^{3/2}}+\frac{1}{x^2}\right) \, dx = 2-\frac{\sqrt{n}+1}{n}\tag{*}$$
Since $\lim_{n\to \infty } \, \text{Ci}\left(\sqrt{n}\right)= 0$ the limit of the partial sum is given by
$$s = \cos(1) - 2 \text{Ci}(1) + \lim_{n\to \infty } \,i_2$$
Observing (*) $s$ remains finite as $n\to\infty$ hence the original sum is convergent. QED.
Remark
A more sophisticated study if the integral $i_2$ might provide better numerical bounds, and even a closed form.
A: We show that  $N\to\sum_{n = 1}^N {\frac{\cos(\sqrt{n})}{n}}$ is a Cauchy sequence and therefore the given series is convergent.
Hint. By using the MVT prove that for $n\geq 1$ and for all $x\in [n,n+1)$
$$\left|\frac{\cos(\sqrt{n})}{n}-\frac{\cos(\sqrt{x})}{x}\right|\leq \frac{1}{n^{3/2}}.$$
Then for $M>N\geq 1$,
$$\begin{align}
&\left|\sum_{n = N}^M {\frac{\cos(\sqrt{n})}{n}}  - \int_N^{M + 1} {\frac{{\cos (\sqrt{x} )}}{x}dx} \right|\\
&=\left|\sum_{n = N}^M {\frac{\cos(\sqrt{n})}{n}}  - \sum_{n = N}^M {\int_n^{n + 1} {\frac{{\cos (\sqrt{x} )}}{x}dx} } \right|\\
&\leq\sum_{n = N}^M \int_n^{n + 1}\left|{\frac{\cos(\sqrt{n})}{n}}  -  { {\frac{{\cos (\sqrt{x} )}}{x}} } \right|dx \le \sum_{n = N}^M \frac{1}{n^{3/2}}.\end{align}$$
Note that since $\sum_{n=1}^{\infty}\frac{1}{n^{3/2}}$ is convergent then
$$\lim_{M,N\to +\infty} \sum_{n = N}^M \frac{1}{n^{3/2}}=0.$$
Moreover
$$\int_1^{+\infty}\frac{{\cos (\sqrt{x} )}}{x}dx=2\int_1^{+\infty}\frac{{\cos (u )}}{u}\, du$$
where the last integral is convergent, and we have that
$$\lim_{M,N\to +\infty}\int_N^{M+1}\frac{{\cos (\sqrt{x} )}}{x}dx=0.$$
A: This approach works in two steps and is generally applicable: first it transforms the sum to an alternating sum which then, in the second step, can be studied by the Dirichlet criterion.
Step 1: Transformation to an alternating sum
The partial sum in question is
$$s_n = \sum_{k=1}^n c_k\tag{1}$$
where
$$c_k = \frac{\cos(\sqrt{k})}{k}\tag{2}$$
Now collecting all subsequent summands with the same sign we can write
$$s_n = \sum_{m=0}^M (-1)^m f_m\tag{3}$$
where
$$f_0 =  \sum_{k=1}^{\lfloor z_{1}\rfloor} c_{k}\tag{4a}$$
$$f_{m\ge 1} = (-1)^m \sum_{k={\lceil z_{m}\rceil}}^{\lfloor z_{m+1} \rfloor} c_{k}\tag{4b}$$
is the sum of terms between two adjacent roots $k=z_m$ and $k=z_{m+1}$ of $c(k) = 0$, and $M$ is some number depending on $n$ which can in principle be specified but it is normally not necessary as it goes to $\infty$ together with $n$
Then, by an appropirate choice of the factor $(-1)^m$,  all $f_m$ can be chosen to be positive quantities.
Step 2: Application of Dirichlet's criterion
Now we can apply the Dirichlet criterion to (3). This reads now as follows: $s_n$ is convergent if
(1) $f_m \to 0$ for $m\to\infty$
(2) $f_m$ is monotonous
ad (1)
We have for $m\ge1$
$$|f_{m}| \le  g(m)$$
where we have dropped the $\cos$ and have defined 
$$g(m) = \sum_{k={\lceil z_{m}\rceil}}^{\lfloor z_{m+1} \rfloor} \frac{1}{k} 
=H_{\lfloor z_{m+1} \rfloor}-H_{\lceil z_{m}\rceil-1}\tag{5} $$
Here $H_n$ ist the harmonic number. 
Now the zeroes of $c_{k}$ are given by
$$z_m = \left(\pi(m-\frac{1}{2})\right)^2\tag{6}$$
Dropping Floor and Ceiling in $g(m)$ defines
$$g_0(m) = H_{z_{m+1}}-H_{z_{m}-1}$$ 
Letting ${\lfloor x \rfloor} \to x-1,{\lceil x \rceil} \to x+1 $ defines another function
$$g_1(m) = H_{z_{m+1}-1}-H_{z_{m}+1-1}$$
and we have the inequality
$$g_1(m) \lt g(m) \lt g_0(m)$$
Asymptotically this leads, up to $O(\frac{1}{m^2})$, to the inequality
$$\frac{2}{m}- \frac{1}{\pi^2 m^2}< g(m) <\frac{2}{m}+\frac{1}{\pi^2 m^2} $$
This shows that $f_m$ goes to zero.
ad (2)
(to be continued)
A: Outline
I haven't fully thought out this answer but leaving it here as a response to a comment on a duplicate question.
At the cost of introducing a constant $C_N$ we can start taking the sum at a large index $N$. $\sum_{n=1} \cos (\sqrt n) = C_N +\sum_{n=N} \cos (\sqrt n)$
Define $x_n = \sqrt n$ so have $\sum_{n=N} \cos (\sqrt n) / n = \sum_{n=N} \cos (x_n) / x_n^2$.
Defining $\Delta x_n = x_{n+1} - x_{n}$ we have $\Delta x_n \approx \frac 1 {2 \sqrt n} = \frac 1 {2 x_n}$ via first order taylor expansion of $x(n+1) \approx x(n) + x'(n)\cdot 1$
Substituting into the sum, we have $ \sum_{n=N} \cos (\sqrt n) / n \approx 2 \sum_{n=N}\cos (x_n) / x_n \Delta x_n$. The right hand side is a Riemann / Darboux integral for $2 \int_{\sqrt N} \frac {\cos x} x dx$ and the integral is finite.
