Can the monotonicity criterion be dropped from the Abel's Test for Uniform convergence?

Question

(Abel's Test) Let $\left(f_n\right)$ and $\left(g_n\right)$ be two sequences of real-valued functions on a set $X$. Assume that: (i) $\sum_n f_n$ is uniformly convergent on $X$. (ii) There exists $M>0$ such that $\left|g_n(x)\right| \leq M$ for all $x \in X$ and $n \in \mathbb{N}$. (iii) $\left(g_n(x)\right)$ is monotone for each $x \in X$. Then the series $\sum_{n=1}^{\infty} f_n g_n$ is uniformly convergent on $X$.

I now present a proof which doesn't require the monotonicity

proof:

$\left|\sum_{k=m+1}^n f_k g_k\right|\leq M\left|\sum_{k=m+1}^n f_k\right|\leq M \cdot \varepsilon / M=\varepsilon .$

This follows from (i) and (ii) of the Theorem. QED

What is the mistake here and what would be a correct proof including the monotonicity criterion?

Answer 1

The error lies in the inequality$$\left|\sum_{k=n+1}^mf_kg_k\right|\leqslant M\left|\sum_{k=n+1}^mf_k\right|.$$All you can deduce from those assumptions is that$$\sum_{k=n+1}^m\left|f_kg_k\right|\leqslant M\sum_{k=n+1}^m\left|f_k\right|,$$which is not the same thing.

Answer 2

The first inequality is not true in general. Let $X$ only conatain one element $x$ and $f_n(x)=\frac{(-1)^n}{n}$, $g_n(x)=(-1)^n$. Then the left side is a divergent series but the right side is bounded.

remark: The other answers were faster, but I still decided to contribute a counterexample.