2
$\begingroup$

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

$\endgroup$
1
  • 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.