I am having trouble in proving the pointwise convergence of $$ g(x)=\sum_{n=1}^\infty \frac{\sin(\sqrt{n}x)}{n}$$ for all real numbers $x$ using elementary methods (e.g. integral test, Weierstrass M-test, Abel's Test, Dirichlet's Test, Comparison with Riemann sum, etc.).
Can someone help me on this?
Best Answer
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.