Convergence and differentiability of a power series

Question

I want to show that

$$\sum_{n=1}^{\infty} \left(x-n \sin\left(\frac{x}{n}\right)\right) $$

converges uniformly and is differentiable on any bounded open interval $(a,b)$ in $\mathbb{R}$.

I don't know how to show uniform convergence here. $ |x- n \sin(\frac{x}{n})| \leq |x-n| $, but that does not provide any help. Do I use the Weierstrass $M$ test? or does it involve the Cauchy criterion?

Regarding the differentiability part, I invoked the taylor series expansion of $\sin{x}$:

$$ \sum_{n=1}^{\infty} \left(x-n \sin\left(\frac{x}{n}\right)\right) = \sum_{n=1}^{\infty} \left(x-n \sum_{m=0}^{\infty} (-1)^{m} \frac{x^{2m+1}}{(2m+1)!} \right)$$

So, can I conclude it is differentiable because the series on the right is differentiable?

Answer 1

Hint

For the uniform converge, make a formal proof based on following arguments:

• For $y$ small enough we have $\sin y \simeq y -\frac{y^3}{6}$.
• Therefore $\vert y - \sin y \vert \le \vert y \vert^3$ for $y$ small enough using a Taylor formula.
• Replacing $y$ by $x/n$ and multiplying by $n$ we get $\vert x -n\sin x/n \vert \le \frac{\vert x \vert^3}{n^2} \le \frac{\max( \vert a \vert^3, \vert b\vert^3)}{n^2}$ for $n$ large enough.
• This allows to conclude to the uniform converge on $[a,b]$.

Regarding differentiability, use a similar approach and a standard theorem about differentiability of series of maps.