This may be very closely related to Robert's suggestion.
Elaborated answer
Consider the following inequality (with $n,m \to \infty$):
$$
\Bigg|\sum\limits_{k = n}^m {\frac{{\sin (\sqrt k x)}}{k}} - \sum\limits_{k = n}^m {\int_k^{k + 1} {\frac{{\sin (\sqrt u x)}}{u}du} } \Bigg| \le \sum\limits_{k = n}^m {\int_k^{k + 1} \Bigg| \frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg|du} .
$$
For fixed $x > 0$, show that, for any $u \in [k,k+1]$,
$$
{\Bigg|\frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg|} \leq \frac{x}{{2k\sqrt k }} + \frac{1}{{k^2 }}.
$$
(Thus the same inequality holds for the integral from $k$ to $k+1$ of the left-hand side.)
For this purpose, first write
$$
\Bigg|\frac{{\sin (\sqrt k x)}}{k} - \frac{{\sin (\sqrt u x)}}{u}\Bigg| = \Bigg|\frac{{\sin (\sqrt k x)k - \sin (\sqrt u x)k + \sin (\sqrt k x)(u - k)}}{{ku}}\Bigg|.
$$
Then apply the triangle inequality, and use the mean value theorem (twice).
Further note that $\int_1^\infty {\frac{{\sin (\sqrt u x)}}{u}} du$ converges (for this purpose, make a change of variable $y=\sqrt u x$). The rest is straightforward.
$\sin (n+\frac{1}{n}) = \sin n \cos \frac{1}{n} + \sin \frac{1}{n} \cos n$
Firstly $\sum_{n=2}^\infty \frac{\sin \frac{1}{n} \cos n}{\ln ^2 n}$ is absolutely convergent from comparison test, because
$\sum_{n=2}^\infty \left| \frac{\sin \frac{1}{n} \cos n}{\ln ^2 n} \right| \le \sum_{n=2}^\infty \frac{\frac{1}{n} \cdot 1}{\ln ^2 n} < \infty$
Convergence of last series follows for example from http://en.wikipedia.org/wiki/Cauchy%27s_condensation_test
Secondly observe, that function $f(x)=\frac{\cos \frac{1}{x}}{\ln ^2 x}$ is decraising for big $x$ because $f'(x)= \frac{\ln x \sin \left( \frac{1}{x} \right) -2x \cos \left( \frac{1}{x} \right)}{\ln ^3 x}$,
Easy to see that $\lim_{x\to \infty} f'(x) = -\infty$ therefore there is $M>0$ such that for $\mathbb{N} \ni n> M$ sequence $\left( f(n) \right)_{n>M}$ is decraising and tends to $0$. If we could show that $\left|\sum_{k=1}^n \sin k \right| $ is bounded, we would just apply Dirichlet test to show that $\sum_{n=2}^\infty \frac{\sin n \cdot \cos \frac{1}{n}}{\ln ^2 n}$ diverges.
It's well known that $\sum_{k=1}^n \sin ka = \frac{\sin \left(\frac{na}{2} \right) \sin \left( \frac{(n+1)a}{2} \right)}{\sin \left( \frac{a}{2}\right)}$ (you can prove it f.e by complex numbers), so
$\left| \sum_{k=1}^n \sin k \right| \le \frac{1}{\left| \sin \frac{1}{2}\right|}$.
To sum up: $\sum_{n=2}^\infty \frac{\sin (n+ \frac{1}{n})}{\ln ^2 n}$ converges as sum of two convergent series.
Best Answer
This is a Clausen function of order 1/2 given by
$$\operatorname{Si}_{1/2}(1)=\sum_{n=1}^\infty\frac{\sin(n)}{\sqrt n}$$
By applying Euler's formula for $\sin(z)=\frac1{2i}(e^{iz}-e^{-iz})$ one gets the relation to the polylogarithm:
$$\operatorname{Si}_{1/2}(1)=\frac1{2i}(\operatorname{Li}_{1/2}(e^i)-\operatorname{Li}_{1/2}(e{-i}i))$$
Quicker numerical evaluation can be done using various relations such as the above. Applying a more brute force approach, one may take the Cesaro summation:
$$\operatorname{Si}_{1/2}(1)=\lim_{n\to\infty}\frac1n\sum_{k=1}^n\sum_{j=1}^k\frac{\sin(j)}{\sqrt j}$$
Evaluating the above at $n=33208$ gives us
$$\operatorname{Si}_{1/2}(1)\simeq1.0439773797177627$$