Pointwise convergence of series $\sum_{k=1}^{\infty}({x\over x+1}\sin(x))^k\quad x\in[0,1]$

Consider the following series $$\sum_{k=1}^{\infty}({x\over x+1}\sin(x))^k,\quad x\in[0,1]$$

I want to investigate the following basic questions: Whether the series converges pointwise on $$[0,1]$$. If yes, does it converge to a continuous function or a discontinuous function? If it converge to continuous function then is the convergence uniform?

Efforts

Let $$F(x)=\sum_{k=1}^{\infty}({x\over x+1}\sin(x))^k,\quad x\in[0,1]$$

$${x\over x+1}\leq{1\over2}$$

$$\sin(x)<1$$

$$\therefore \left({x\over x+1}\sin(x)\right)\leq {1\over 2}$$

A series of functions $$\sum f_n$$ will converge uniformly on $$[a,b]$$ if there exist a convergent series $$\sum M_n$$ of positive numbers such that for all $$x\in [a,b]$$ $$|f_n(x)|\leq M_n\quad \forall n$$

I think we can consider the series $$\sum_{k=1}^{\infty}({1\over 2})^k$$

and hence we can conclude that series converges uniformly on $$[0,1]$$.

Is my procedure correct? To what function does it converges. I think if we consider is as geometric series with $$r={x\over x+1}\sin(x)$$

So it will converges to $$F(x)={\frac{x}{x+1}\sin(x)\over 1-\frac{x}{x+1}\sin(x)}$$

• The procedure in general looks right, and I'd write $$F(x)=\frac{x\sin x}{x(1-\sin x)+1}$$ – DonAntonio Oct 27 '18 at 13:12

This looks quite correct to me and, yes, it is just a geometric series so that is the sum you will get. You can make it a bit prettier by simplifying: $$F(x)=\frac{x\sin x}{x+1-x\sin x}$$