# $x+\sum _{i=2}^{i=n}\left(x^{\left(2i-1\right) ^{-1}}-x^{\left( 2i-3\right) ^{-1}}\right) = ?$

I am trying to show that near $x = 0$ the series $u_1(x) +u_2(x) +u_3(x)+\dots$, where $u_1(x) = x$, and $u_n(x) = x^{\left(2n-1\right) ^{-1}}-x^{\left( 2n-3\right) ^{-1}}$, and real values of $x$ are concerned , is discontinuous and non uniformly convergent.

Now of course the necessary condition for uniformity of convergence is that. If $R_{n,p}\left( x\right) = u_{n+1}\left( x\right) +u_{n+2}\left( x\right)+\dots+ u_{n+p}\left( x\right)$ given any positive number $\epsilon$, it should be possible tp choose $N$ independent of x(but depending on $\epsilon$ such that $$\left|R_{n,p}(x)\right|<\epsilon$$ for all positive integral values of $p$. $$\left| S\left( x\right) -S_{n}\left( x\right) \right| = \left| \lim _{p\rightarrow \infty }R_{n,p}\left( x\right) -R_{N,n-N}\left( x\right) \right| < \epsilon$$ whenever $n>N$ and N is independent of x.

I was hoping to show that given an arbitrarily small $\epsilon$, it is possible to choose values of $x$, as small as we please, depending on $n$ in such a way that $\left|R_{n}(x)\right|$ is not less than $\epsilon$ for any value of $n$, no matter how large. The existence of such values of $x$ is inconsistent with the condition for uniformity of convergence, though i am having a a hard time finding a $S_n(x)$ and $R_n(x)$. would this be a valid argument ?

Any help or alternative proof strategies would be much appreciated.

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Note that $$\sum\limits_{i=2}^n(x^{\frac{1}{2i-1}}-x^{\frac{1}{2i-3}})= \sum\limits_{i=2}^n x^{\frac{1}{2i-1}}- \sum\limits_{i=2}^n x^{\frac{1}{2i-3}}= \sum\limits_{i=2}^n x^{\frac{1}{2i-1}}- \sum\limits_{i=1}^{n-1} x^{\frac{1}{2i-1}}=$$ $$\left(\sum\limits_{i=2}^{n-1} x^{\frac{1}{2i-1}}+x^{\frac{1}{2n-1}}\right)- \left(x^{\frac{1}{1}}+\sum\limits_{i=2}^{n-1} x^{\frac{1}{2i-1}}\right)= x^{\frac{1}{2n-1}}-x$$ Hence $$x+\sum\limits_{i=2}^n(x^{\frac{1}{2i-1}}-x^{\frac{1}{2i-3}})=x^{\frac{1}{2n-1}}$$ It is remains to notice that $$S(0)=\sum\limits_{i=1}^\infty u(0)=\lim\limits_{n\to\infty} 0^{\frac{1}{2n-1}}=0$$ $$S(x)=\sum\limits_{i=1}^\infty u(x)=\lim\limits_{n\to\infty} x^{\frac{1}{2n-1}}=1\qquad 0<x\leq 1$$ i.e. $S$ is not continuous, because $$\lim\limits_{x\to 0+} S(x)\neq S(0)$$ Our series consist of continuous functions. If it converges uniformly then $S(x)$ must be continuous too. Contradiction.