Pointwise Convergence of $\sum \frac{\sin(\sqrt{n}x)}{n}$

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.

Look at the sequence of partial sums up to each sign change in the sine, i.e. combine all consecutive terms with the same sign. Since the length of these groups of terms grows as $\sqrt{n}$, their sum decays as $1/\sqrt{n}$, so you get an alternating series with terms converging to zero.

Edit: Didier rightly pointed out that I didn't show that the absolute values of the terms strictly decrease. To show this rigorously would probably be more hassle than pursuing one of the approaches in the other answers, but here's a not-so-rigorous argument: Each group of terms can be considered as an approximation with the trapezoidal rule of the integral over $\sin \sqrt{ux}/u$ over an interval of $\pi/x$, scaled in proportion to the number of terms in the group. The scaling by the number of terms grows as $\sqrt{n}$ and the integrand falls off as $1/n$, for a decay of $1/\sqrt{n}$. There are three sources of approximation error. Two occur at the endpoints, one from the fact that the trapezoidal rule specifies a weight of $1/2$ there, and another from the fact that the ends of the sum and the integral don't coincide. Both of these errors are proportional to the outermost terms in the group, which are of order $1/\sqrt{n}^3$, and to the width of the integration interval they represent, which is of order $1/\sqrt{n}$, for a total decay of $1/n^2$. The third approximation error is the error in the trapezoidal rule itself, which is cubic in the width of the integration intervals for a factor of $1/\sqrt{n}^3$, and proportional to the second derivative of the integrand, which falls off with $1/n$, for a total decay of $1/\sqrt{n}^5$. So both approximation errors decay faster than the difference of order $1/\sqrt{n}^3$ in the contributions of order $1/\sqrt{n}$.

You can use summation by parts here: If $(a_n)_{n \geq 1}$ is a sequence of complex numbers, and $f : [1, \infty] \longrightarrow {\bf C}$ is a $C^1$ function then $$ \sum_{n \leq N} a_n f(n) = (\sum_{n \leq N} a_n) f(N) - \int_{1}^{N} (\sum_{n \leq t} a_n) f'(t) \; d \, t.$$ Applying this to $a_n = {1 \over n}$ and $f(t) = \sin(\sqrt{t}x)$, one gets $$\sum_{n \leq N} {\sin(\sqrt{n}x) \over n} = H(N)\sin(\sqrt{N}x) - \int_{1}^{N} H(\lfloor t \rfloor) {x\cos(\sqrt{t}x) \over 2\sqrt{t}} \; d \, t$$ Here $H(n)$ is the $n$th harmonic number $1 + {1 \over 2} + ... + {1 \over n}$. Fortunately, there are good asymptotics for $H(n)$, namely that $H(n) = \ln(n) + \gamma + O(1/n)$. As a result, the $H(\lfloor t \rfloor)$ appearing in the above integral differs from $\ln(t) + \gamma$ by at most ${C \over t}$. Since ${\displaystyle \int_1^{\infty} {1 \over t} {x\cos(\sqrt{t}x) \over 2\sqrt{t}}\,dt}$ is absolutely integrable, one can replace $H(\lfloor t \rfloor)$ by $\ln(t) + \gamma$ in that integral when trying to prove convergence. One can similarly disregard the $O({1 \over N})$ term in the $H(N)$ on the left. So our goal is to show that the following converges as $N$ goes to infinity. $$(\ln(N) + \gamma) \sin(\sqrt{N}x) - \int_{1}^{N} (\ln(t) + \gamma){x\cos(\sqrt{t}x) \over 2\sqrt{t}} \; d \, t.$$ We now integrate by parts again, in the opposite direction of our original summation by parts; we differentiate $(\ln(t) + \gamma)$ and integrate ${x\cos(\sqrt{t}x) \over 2\sqrt{t}}$ back to $\sin(\sqrt{t} x)$. Our goal now becomes to show the finiteness of $$\int_1^{\infty}{\sin(\sqrt{t} x) \over t}\,dt$$ As others have said above, by changing $t$ to $t^2$, this is equivalent to showing the convergence of ${\displaystyle \int_1^{\infty} {\sin(t x) \over t}\,dt}$, which in turn by replacing $t$ by $tx$ is equivlalent to showing the convergence of ${\displaystyle \int_x^{\infty} {\sin(t) \over t}\,dt}$ which is well-known.