prove or disprove $\sum_{n=1}^\infty \frac{(-1)^{n+1}}{x^2+2n- \sin nx}$ converge uniformly in $\mathbb{R}$

Question

I need to prove or disprove that $$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{x^2+2n- \sin nx}$$ converge uniformly in $\mathbb{R}$. First I tried using Weierstrass theorem to prove that the series converge uniformly, but I couldn't find a suitable series, so I moved to try to both prove and disprove uniform convergence using Cauchy criterion, but I'm was unable to neither prove that the criterion is fulfilled for every $\varepsilon > 0$ nor find an $\varepsilon$ that disprove it

Answer 1

Weirestrass's test is too weak here, as your series doesn't converge absolutely.

First, lets get rid of $\sin$: say $$\sum\limits_{n=1}^\infty \frac{(-1)^{n + 1}}{x^2 + 2n - \sin nx} =\\ \sum\limits_{n=1}^\infty\frac{(-1)^{n+1} (-\sin nx)}{2n(x^2+2n-\sin nx)} + \sum\limits_{n=1}^\infty \frac{(-1)^{n+1}}{x^2 + 2n} =\\ \sum\limits_{n=1}^\infty\frac{(-1)^{n+1} (-\sin nx)}{2n(x^2+2n-\sin nx)} + \sum\limits_{n=1}^\infty\frac{(-1)^{n+1}}{2n} - \sum\limits_{n=1}^\infty \frac{(-1)^{n+1} x^2}{2n(x^2 + 2n)} $$.

The first series converges uniformly by Weierstrass's test (for $n > 2$ common term is bounded by $\frac{1}{n^2}$).

The second converge uniformly as it converges and doesn't depend on $x$ at all

The third converges uniformly by Abel's test, take $a_n = \frac{(-1)^{n+1}}{n}$ and $f_n(x) = \frac{x^2}{x^2 + 2n}$.