<s>[INST] Uniform convergence of $\sum\limits_{k = 1}^{\infty} \frac{\sin (\sqrt{x}/k)}{\sqrt{x^2 + k^2}}$ I need to check uniform convergence of $\sum\limits_{k = 1}^{\infty} \frac{\sin (\sqrt{x}/k)}{\sqrt{x^2 + k^2}}, x \in [0, +\infty)$. I've tried Weierstrass Test, but it didn't work.
 [/INST] The convergence is not uniform for $x \in [0,\infty)$. 
With $x_n = \frac{\pi^2n^2}{4}$ and $n < k \leqslant 2n$ we have 
$$\frac{\pi}{4} = \frac{\pi n}{2} \frac{1}{2n} \leqslant \frac{\sqrt{x_n}}{k} < \frac{\pi n}{2} \frac{1}{n}=\frac{\pi}{2}$$
Thus, for all $n \in \mathbb{N}$
$$\left|\sum_{k=n+1}^{2n}\frac{\sin \frac{\sqrt{x_n}}{k}}{\sqrt{x_n^2 + k^2}} \right| =\sum_{k=n+1}^{2n}\frac{\sin \frac{\sqrt{x_n}}{k}}{\sqrt{x_n^2 + k^2}} \geqslant n \cdot\frac{ \frac{1}{\sqrt{2}}}{\sqrt{\frac{\pi^2}{4}n^2 + 4n^2}}\\ = \frac{1}{\sqrt{\frac{\pi^2}{2} + 8}}$$
This violates the Cauchy criterion for uniform convergence which requires that for any $\epsilon > 0$ there exists a positive integer $N$ such that for all $m > n > N$ and for all $x \in [0,\infty)$ we have
$$\left|\sum_{k=n+1}^{m}\frac{\sin \frac{\sqrt{x}}{k}}{\sqrt{x^2 + k^2}} \right| < \epsilon$$ 
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