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?
 A: 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.
A: Note that
$$
\sqrt{k+1}-\sqrt{k}=\frac1{\sqrt{k+1}+\sqrt{k}}\le\frac1{2\sqrt{k}}\le\frac1{\sqrt{k}+\sqrt{k-1}}=\sqrt{k}-\sqrt{k-1}
$$
Therefore,
$$
2\left(\sqrt{n+1}-\sqrt{m}\right)\le\sum_{k=m}^n\frac1{\sqrt{k}}\le2\left(\sqrt{n}-\sqrt{m-1}\right)
$$
No matter how big $m$ is, you can't bound the sum for all $n$.
Your argument seems to indicate that for all $\varepsilon\gt0$, there is an $m$ so big that
$$
\sum_{k=m}^n\frac1{\sqrt{k}}\le\varepsilon
$$
for all $n\ge m$.

Now that I can read your answer better, the problem with your answer is that $p$ should not be dependent on $n$. That is, for any $\varepsilon\gt0$, there must be an $N$ (dependent on $\varepsilon$) so that for any $n\ge N$ and any $p\ge0$,
$$
\left|\sum_{k=n}^{n+p}\frac{x^k}{\sqrt{k}}\right|\le\varepsilon
$$

Let me expand on what I have said in the last section.
What you have shown that for a fixed $p$,
$$
\lim_{n\to\infty}\sum_{k=n+1}^{n+p}\frac1{\sqrt{k}}=0
$$
and this is true. However, this is not sufficient to show that
$$
s_n=\sum_{k=1}^n\frac1{\sqrt{k}}
$$
is a Cauchy sequence. To show that $s_n$ is a Cauchy sequence, you need to show that for any $\varepsilon\gt0$, there is an $M$ so that for all $n\ge m\ge M$
$$
\begin{align}
\left|s_n-s_m\right|
&=\left|\sum_{k=m+1}^n\frac1{\sqrt{k}}\right|\\
&\le\varepsilon
\end{align}
$$
Your proof only shows this for $n-m\le p$.
