Cauchy Criterion - Prove $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{\sqrt k}$ converges uniformly on $[0,1]$ $$\sum_{k=1}^{\infty} (-1)^{k-1}\frac{x^{k}}{\sqrt k}$$
I have shown this converges uniformly in $[0,1]$:
But my working also implies $\sum_{k=1}^{\infty}\frac{x^{k}}{\sqrt k}$ converges uniformly in $[0,1]$ but $\sum_{k=1}^{\infty}\frac{1}{\sqrt k}$ clearly diverges. 
Define $S_n(x) = \sum_{k=1}^{n}\frac{x^{k}}{\sqrt k} \forall x\in [0,1]$
I will show $S_n$ is uniformly cauchy on $[0,1]$
Now $\lim_{n\to\infty} \frac{p}{\sqrt n}=0$ so using same argument as before $\exists N s.t. \forall n>N, \left|{\frac{1}{\sqrt n}}\right|<\frac{\epsilon}{p} $
Then $\forall x \in [0,1], \forall p$ and $\forall n>N$ we have:
$\left|s_{n+p}(x)-s_n(x)\right| = \left|\sum_{k=n+1}^{n+p}\frac{x^{k}}{\sqrt k}\right| ≤  \left|\sum_{k=n+1}^{n+p}\frac{1}{\sqrt k}\right| ≤\left|{\frac{1}{\sqrt n}}\right|+...+\left|{\frac{1}{\sqrt n}}\right|<p.\frac{\epsilon}{p} =\epsilon$
Where is my mistake?

To show that $S_n(x)$ is uniformly Cauchy on $[0,1]$, you need to find $N(\epsilon)$ (depending only on $\epsilon$) such that regardless of $p \ge 0$,
$$n > N \implies \sum_{n}^{n+p} < \epsilon$$
However, "your $N$", as it is now, depends on $p$; i.e, the bigger $p$ is, so does your $N$ have to be.
