I need to find if the series $$\sum_{k=2}^\infty\left(\cos\frac{x}{k}-\cos\frac{x}{k-1}\right)$$ converges uniformly on $(-\infty,\infty)$.
My attempt:
The partial sum of the series: $$\sum_{k=2}^n\left(\cos\frac{x}{k}-\cos\frac{x}{k-1}\right)=\cos\frac{x}{n}-\cos{x}.$$ Therefore, the sum of the series is $$S(x)=1-\cos x.$$ I know that I now need to show that $\cos\frac{x}{k}-\cos\frac{x}{k-1}$ uniformly converges to $1-\cos x$ on $(-\infty,\infty)$. Unfortunately, I don't know how to do that. My guess would be to use Cauchy criterion for uniform convergence but I haven't been really successful on that front. Any help is appreciated!
The convergence is not uniform. If it was, there would be some $\require{cancel}N\in\Bbb N$ such that$$(\forall x\in\Bbb R)(\forall n\in\Bbb N):n\geqslant N\implies\left|(1-\cancel{\cos(x)})-\left(\cos\left(\frac xn\right)-\cancel{\cos(x)}\right)\right|<1,$$which is equivalent to$$(\forall x\in\Bbb R)(\forall n\in\Bbb N):n\geqslant N\implies\left|1-\cos\left(\frac xn\right)\right|<1.$$In particular$$(\forall x\in\Bbb R):\left|1-\cos\left(\frac xN\right)\right|<1.$$But if $x=\pi N$, then $\left|1-\cos\left(\frac xN\right)\right|=2$.