If the sequence of functions $f_{n}: X \longrightarrow \mathbb{R}$ is such that $f_{1} \geq ... \geq f_{k} \geq ...$ and $f_{n} \longrightarrow 0$ uniformly in $X$. Prove that $\displaystyle \sum(-1)^{n}f_{n}$ uniformly converges in $X$
Since $f_{n}$ is uniformly convergent, the convergence of $f$ not dependes of $x$. Thus, is the proof reduced to the Leibniz test?
In particular, $f_n\to 0$ pointwise and $f_n\geq 0$, so $S(x):=\sum_{n=1}^\infty (-1)^nf_n(x)$ can be defined as a function by the alternating series test. Moreover, the truncation error for only taking $N$ terms is bounded by the $N+1$th term. This can be seen as follows: let $S_N$ be the $N$th partial sum, and let $g_n = f_{2n} - f_{2n-1}$. Then $g_n≤0$, so $S_{2N} = \sum_{n=1}^N g_n$ must decrease to $S$. Similarly, $S_{2N+1}$ must increase towards $S$. Therefore, for any $N$, $$ S_{2N+1} ≤ S_{2N+3} ≤ S ≤ S_{2N+2} ≤ S_{2N}$$ and it follows (by $f_n$'s uniform convergence) that $$ |S(x) - S_{N}(x)| ≤ |f_{N+1}(x)| ≤ \|f_{N+1}\|_{\infty,X} \to 0.$$ Hence, $S_N$ converges uniformly to $S$.