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?


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

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


$\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 $$F(x)={\frac{x}{x+1}\sin(x)\over 1-\frac{x}{x+1}\sin(x)}$$

  • 1
    $\begingroup$ The procedure in general looks right, and I'd write $$F(x)=\frac{x\sin x}{x(1-\sin x)+1}$$ $\endgroup$ – 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: $$ F(x)=\frac{x\sin x}{x+1-x\sin x} $$


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