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.