Testing a series for uniform convergence using Weierstrass' M test

Question

I'm currently having some trouble trying to test for uniform convergence of the series.

$\sum_{k=0}^{\infty}\frac{1}{kx+2}-\frac{1}{kx+x+2}$ $0 \leq x \leq 1 $

I tried to test for uniform convergence using the Weierstrass' M test where I set my M such that $$ M_k=\frac{1}{k+2} $$ and $$\sum_{k=0}^{\infty}\frac{1}{k+2} $$

After performing a comparison test, I came to the conclusion that $\sum_{k=0}^{\infty}\frac{1}{k+2} $ has divergence.

I understand that suppose $|u_k| \leq M_k $, if $$\sum_{k=0} M_k < \infty$$ than $$\sum_{k=0} u_k $$ converges uniformly in a $\leq x \leq b$.

However, I haven't been able to find or understand, what if $$\sum_{k=0}^{\infty} M_k$$ divergence..does this imply that $$\sum_{k=0}^{\infty} u_k$$ diverges too? or is there an alternative method to prove that a series does not uniformly converge but converges?

Answer 1

To clarify your confusion:

Weierstrass M-test states one and only one thing:

Given a sequence of functions $f_k(x)$ defined on $E\subseteq\mathbb{R}$, the series $\sum f_k(x)$ converges uniformly if there exists a sequnece of reals $M_k$ such that $|f_k|\le M_k$ for each $k$ and $\sum M_k$ converges.

Note here that $M_k$ must not depend on $x$. This only means that if you found such sequence $M_k$ then the series uniformly converges. It says nothing about the series if you have found some $M_k$ whose series does not converge.

In particular, this test cannot be used to prove that some series does not converge uniformly.

Now consider

$$ f_k(x)=\frac{1}{kx+2}-\frac{1}{(k+1)x+2} $$

Consider the partial sum

\begin{align*} S_n(x)=\sum_{k=0}^{n} f_k(x)&=\left(\frac{1}{2}-\frac{1}{x+2}\right)+\cdots+\left(\frac{1}{nx+2}-\frac{1}{(n+1)x+2}\right)\\ &=\frac{1}{2}-\frac{1}{(n+1)x+2} \end{align*}

the series $\sum f_k(x)$ converges uniformly if and only if $S_n$ converges uniformly.

Now $S_n(0)=0$ for all $n$. So $S_n(0) \to 0$, and $S_n(x) \to \frac{1}{2}$ for $0<x\le 1$. So define

$$ S(x)= \begin{cases} 0 & \text{if} \quad x=0 \\ \frac{1}{2} & \text{if} \quad 0<x\le 1 \end{cases} $$

Then $S_n(x)\to S(x)$. Also, $S_n(0)-S(0)=0$ for any $n$. Now

$$ m_n:=\sup_{x\in [0,1]} |S_n-S|=\sup_{x\in (0,1]} \left\lvert -\frac{1}{(n+1)x+2} \right\rvert=\frac{1}{2} \not\to 0 $$

So $S_n$ does not converge uniformly.

Answer 2

For all $x\in(0,1)$, $$ \begin{align} \sum_{k=0}^n\left(\frac1{kx+2}-\frac1{kx+x+2}\right) &=\sum_{k=0}^n\frac1{kx+2}-\sum_{k=0}^n\frac1{kx+x+2}\\ &=\sum_{k=0}^n\frac1{kx+2}-\sum_{k=1}^{n+1}\frac1{kx+2}\\ &=\frac12-\frac1{(n+1)x+2}\tag{1} \end{align} $$ the limit is $\frac12$. However, as $x\to0$, we need $n\ge\frac{1-2\epsilon}{\epsilon x}\to\infty$ for the expression to be within $\epsilon$ of $\frac12$. Thus, the convergence is not uniform on $(0,1)$ so it cannot be uniform on $[0,1]$.

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Another approach is to use the fact that a sequence of continuous functions that converge uniformly, converge to a continuous function. Equation $(1)$ shows that the sum converges to $\frac12$ on $(0,1]$ and $0$ at $0$, which is not continuous on $[0,1]$.

Answer 3

Define the parital sum as $S_n(x)=\sum_{k=0}^{n} f_k(x)$

$$S_n(x)=\left(\frac{1}{2}-\frac{1}{x+2}\right)+\cdots+\left(\frac{1}{nx+2}-\frac{1}{(n+1)x+2}\right)=\frac{1}{2}-\frac{1}{(n+1)x+2}$$

First check pointwise convergence

If $x=0,~~S_n(0)=0$,

If $x\in (0,1]$, $\lim_{n\to\infty} S_n(x)=\frac{1}{2}$.

$$ S(x)= \begin{cases} 0 & \text{if} \quad x=0 \\ \frac{1}{2} & \text{if} \quad 0<x\le 1 \end{cases} $$

Since $S_n(x)$ is continous for all $n$, if $S_n(x)\to S(x)$ uniformly, then $S(x)$ is continuous.

Obviously, $S(x)$ is not continous at $0$, (by $p\to q \Leftrightarrow\neg q\to\neg p $), therefore $S_n$ is not uniformly convergent.