Cauchy Criterion - Prove $\sum_{k=1}^{\infty}(-1)^{k-1}\frac{x^k}{\sqrt k}$ converges uniformly on $[0,1]$

Question

$$\sum_{k=1}^{\infty} (-1)^{k-1}\frac{x^{k}}{\sqrt k}$$

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?

Answer 1

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 $n\ge m$.

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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 $p\ge0$, $$ \left|\sum_{k=n}^{n+p}\frac{x^k}{\sqrt{k}}\right|\le\varepsilon $$

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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 $n-m\le p$.

Answer 2

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.