Show that the series $\sum \frac{\sin(n)}{1+n^{2}x^{2}}$ converges uniformly.

Question

Show that the series $\sum \frac{\sin(n)}{1+n^{2}x^{2}}$ converges uniformly on the set where $x$ is real and $|x|\geq \varepsilon$.

I wanted to do the Weierstrass M-test. Can I just say that $\left | \frac{\sin(n)}{1+n^{2}x^{2}} \right |\leq \left | \frac{1}{1+n^{2}} \right |$, and $\sum \frac{1}{1+n^{2}}$ converges by the comparison test or am I supposed to say something about $x$?

Answer 1

Since it doesn't converge uniformly if we take an interval containing zero - the series at $x=0$ is $\sum_n \sin n$, which doesn't converge at all since its terms don't go to zero - we absolutely need something about $x$ in there.

Indeed, your inequality $\left|\frac{\sin n}{1+n^2x^2}\right| \le \frac1{1+n^2}$ is only true for $|x|\ge 1$. For the domain $|x|\ge\epsilon$ we're working with here, we have to settle for the weaker inequality $$\left|\frac{\sin n}{1+n^2x^2}\right| \le \left|\frac{1}{1+n^2x^2}\right| \le \frac1{1+n^2\epsilon^2}$$ That series converges by limit comparison, and then we can apply the M-test.

Answer 2

Let $\varepsilon>0.$ One may use the inequality $$ \left | \frac{\sin(n)}{1+n^{2}x^{2}} \right |\leq \left | \frac{1}{1+n^{2}x^{2}} \right |\leq\frac{1}{1+\varepsilon^2n^{2}} , \qquad |x|\ge\varepsilon, $$ then concluding with the Weierstrass M-test.