I am self-learning real-analysis from the text, Understanding Analysis, by Stephen Abbott. I would like someone to verify if my below counterexample to the exercise problem 6.4.2 (c) is valid and technically correct.
[Abbott 6.4.2] Decide whether each proposition is true or false, providing a short justification or counterexample as appropriate.
(c) If $\sum_{n=1}^{\infty} f_n$ converges uniformly on $A$, then there exists constants $M_n$ such that $|f_n(x)| \leq M_n$ for all $x \in A$ and $\sum_{n=1}^{\infty}M_n$ converges.
Proof.
The converse statement is:
$(\sum_{n=1}^{\infty} f_n$ converges uniformly on $A)$ $\implies$ $\{[(\exists M_n)(|f_n(x)|\leq M_n)(\forall x \in A$)]$ \land$ $[\sum_{n=1}^{\infty}M_n$ converges$]\}$
Carefully negating this, we have:
$(\sum_{n=1}^{\infty} f_n$ converges uniformly on $A)$ $\land$ $\{\lnot[\cdots]$ $\lor$ $\lnot[\sum_{n=1}^{\infty}M_n$ converges$]\}$
Let $(f_n)$ be the sequence of functions defined by:
$$f_n(x) = \frac{x^n}{n}, \quad x\in\left[0,\frac{1}{2}\right]$$
The infinite series
$$\sum_{n=1}^{\infty}f_n(x)=x+\frac{x^2}{2} + \frac{x^3}{3}\ldots$$
converges uniformly on $\left[0,\frac{1}{2}\right]$. To see this, pick an arbitrary $\epsilon > 0$ and consider:
\begin{align*} |f_{m+1}(x) + \ldots + f_n(x)| &= f_{m+1}(x) + \ldots + f_n(x) & \{f_n(x) \geq 0\}\\ &= \frac{x^{m+1}}{m+1} + \ldots + \frac{x^n}{n} \\ &\leq \frac{x^{m+1}}{m+1} + \frac{x^{m+2}}{m+1} + \ldots + \frac{x^{n}}{m+1}\\ &\leq \frac{1}{m+1}\left[1+x+x^2+\ldots\right]\\ &= \frac{1}{m+1}\cdot\frac{1}{1-x}\\ &\leq \frac{1}{m}\cdot \frac{1}{1-(1/2)}= \frac{2}{m} & \{x \leq \frac{1}{2}\} \end{align*}
Pick $N > \frac{2}{\epsilon}$. Then, $|f_{m+1}(x) + \ldots + f_n(x)| < \epsilon$ for all $n > m \geq N$ and for all $x \in [0,\frac{1}{2}]$. By the Cauchy criterion for the uniform converge of infinite series, $\sum_{n=0}^{\infty}f_n(x)$ is uniformly convergent.
Moreover, pick $M_n = \frac{1}{n}$. We have:
$$f_n(x) = \frac{x^n}{n} \leq \frac{1}{n} = M_n$$
for all $x\in [0,\frac{1}{2}]$. But, $\sum_{n=1}^{\infty}\frac{1}{n}$ is the harmonic series which is divergent.