M-Test for convergence How do I use the M-Test to show that the follow series converges uniformly for $x \in (0, \infty)$. We already know that it is convergent in the domain. 
$$ \sum_{k=1}^{\infty}\frac{-2k}{(1+kx)^{3}}$$ 
I cannot figure out what to compare this to.
 A: I claim that the series that you gave does not in fact converge uniformly for $x \in (0,\infty)$. Consider the case of a generic series
$$s(x) = \sum_{k =1}^\infty a_k(x),$$
and its sequence of partial sums
$$ s_n(x) = \sum_{k =1}^n a_k(x),$$
which we assume satisfies $s_n(x) \rightarrow s(x)$, as $n \rightarrow \infty$. If in addition we have uniform convergence on $(0,\infty)$, then for all $\epsilon > 0$, there is an $N \geq 1$ such that for all $x \in (0,\infty)$
$$ |s_n(x) - s(x)| < \epsilon, \qquad \text{for } n \geq N.$$
In particular this says that for $n \geq N$
$$
\sup_{x \in (0,\infty)} |s_n(x) - s(x)| < \epsilon.
$$
In the following I will contradict this by showing that for the series in question, this supremum is in fact $\infty$, for all $n \geq 1$. We have
\begin{align}
s_n(x) = -\sum_{k=1}^n \frac{2 k}{(1+ k x)^3},
\end{align}
and
\begin{align}
|s_n(x) - s(x)| &= \sum_{k=n+1}^\infty \frac{2k}{(1+k x)^3}\\
& > \frac{2(n+1)}{(1 + (n+1)x)^3},
\end{align}
where we have simply bounded the series by its first term, which we can do since all terms are positive. But then
$$ 
\sup_{x \in (0,\infty)} |s_n(x) - s(x)| \geq \sup_{x \in (0,\infty)} \frac{2(n+1)}{(1 + (n+1)x)^3} = \infty.
$$
Hence for all $n \geq 1$
$$\sup_{x \in (0,\infty)} |s_n(x) - s(x)| = \infty.$$
Edit. Following Hasan Saad's comment, we can show uniform convergence of the series on the domain $[c,\infty)$ for any $c > 0$. To this end, note that the coefficients $a_k(x) = -2k/(1 + k x)^3$ are monotonic increasing, and hence $$ |a_k(x)| \leq |a_k(c)| = \frac{2k}{(1 + k c)^3}, \qquad x \geq c. $$
Now let $M_k = a_k(c)$ be the sequence of bounding constants for Weierstrass' M-test. According to the original post, it is already known that $\sum M_k < \infty$, since it was already known that the series $s(x)$ converged. Alternatively we can prove this directly
\begin{align}
\sum_{k=1}^\infty M_k & = \sum_{k \geq 1} \frac{2k}{(1 + k c)^3} \\
& \leq \sum_{k \geq 1} \frac{2k}{(k c)^3} \\
& \leq \frac{2}{c^3} \sum_{k \geq 1} \frac{1}{k^2} \\
& = \frac{ \pi^2}{3 c^3},
\end{align}
in particular this is finite, so the M-test asserts uniform convergence for $x \in [c,\infty)$.
