Show that $f(x) = \sum_{k=0}^{\infty}{\frac{\sin^2(kx)}{1+k^2 x^2}}$ uniformly converges.

Question

I want to show that $f(x) = \sum_{k=0}^{\infty}{\frac{\sin^2(kx)}{1+k^2 x^2}}$ uniformly converges for $|x| \geq \delta$ for any given $\delta > 0$.

I don't know how to use the M-test here, since it seems each term is constantly upper bounded by 0.359, regardless of the value of $k$. I tried performing the Cauchy Criterion, but I am having a hard time showing that an $N$ exists for all $|x| \geq \delta$.

Any advice or hints would be great! For context, I'm a first year undergraduate maths student.

Answer 1

See that, for every non-negative integer $k$, $$|x| \geq \delta \Rightarrow x^2 \geq \delta^2 \Rightarrow 1 + k^2 x^2 \geq 1 + k^2 \delta^2 \Rightarrow \frac{1}{1 + k^2 x^2} \leq \frac{1}{1 + k^2 \delta^2}.$$ As the absolute value of $\sin$ is bounded by $1$, it follows that $$\Biggl| \frac{\sin^2{(kx)}}{1 + k^2 x^2} \Biggr| \leq \frac{1}{1 + k^2 \delta^2}$$ for every non-negative integer $k$ and for every real $x$ such that $|x| \geq \delta$. But the series $\sum_{k = 0}^{\infty} 1/(1 + k^2 \delta^2)$ converges: using comparison, one sees that, for every positive integer $k$, $$1 + k^2 \delta^2 > k^2 \delta^2 \Rightarrow \frac{1}{1 + k^2 \delta^2} < \frac{1}{k^2 \delta^2},$$ and the series $\sum_{k = 1}^{\infty} 1/(k^2 \delta^2)$ converges, as $\sum_{k = 1}^{\infty} 1/k^2$ converges. Now you can apply the M-test.