I'm trying to prove uniform convergence and differentiability for any bounded open interval ($a,b$) $\in \mathbb{R}$ $$ \sum_{n=1}^{\infty} \left(x-n\sin \left(\frac x{n}\right)\right) $$
I was trying to use Weierstrass M-test to prove this but I cannot find an $M_n$ that is greater than or equal to the absolute value of the inside function for all of $n$.
Any help, alternative methods or suggestions for $M_n$ or anything else, would be greatly appreciated.
On
Let consider derivative of function $f(x)=x-n\sin \left(\frac x{n}\right)$, then we have $f'(x)=1-\cos\frac {x}{n}\geqslant 0$, which gives, that $f$ is increasing and it's supremum for any $(a, b)$ is $f(b)=b-n\sin \left(\frac {b}{n}\right)$.
Knowing $b-n\sin \left(\frac {b}{n}\right) = n \left(\frac {b}{n}-\sin \left(\frac {b}{n}\right)\right) = n \left(\frac {b}{n}- \left[\frac {b}{n}- \frac {1}{6}\left(\frac {b}{n}\right)^3 + O\left(\left(\frac {b}{n}\right)^5\right)\right] \right) =O\left( \frac {1}{n^2}\right)$ we have Weierstrass M-test.
For differentiability we should take look at $ \sum\limits_{n=1}^{\infty} \left(1-\cos\frac {x}{n}\right) = \sum\limits_{n=1}^{\infty} 2\sin^2 \frac {x}{2n}$ which uniform convergence let me leave to you
Let's say your interval is $(0,b)$ (Take $a = 0$), then: $x- n\sin \left(\frac{x}{n}\right)= n\left(\dfrac{x}{n} - \sin\left(\frac{x}{n}\right)\right)\le \dfrac{n}{6}\cdot \left(\dfrac{x}{n}\right)^3 \le \dfrac{b^3}{n^2}$. Thus you might take $M_n = \dfrac{b^3}{6n^2}$. Note here we use the well-known fact: $x - \sin x \le \dfrac{x^3}{6}, x \in (0,b), b $ is small.