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?
Best Answer
To clarify your confusion:
Weierstrass M-test states one and only one thing:
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.