Testing a series for uniform convergence using Weierstrass' M test 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?
 A: 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.
A: 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 $[0,1]$.

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 $[0,1]$.
A: 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.
