Show uniform convergence of function series 
Let the function sequence $\left\{ f_n \right\}_{n\ge n_0}$ satisfy:
(1) $\forall_{x\in D}\forall_{n\in\mathbb{N}} \ f_n(x)\ge 0$
(2) $\forall_{x\in D}\forall_{n\in\mathbb{N}} \ f_n(x)\ge f_{n+1}(x)$
(3) $\sup_{x\in D}f_n(x)\rightarrow 0$
$D$ - domain
Show that function series $\displaystyle\sum_{n=n_0}^{+\infty}(-1)^{n+1}f_n(x)$ converges uniformly.

I don't know how to approach this. Hope the solution isn't too hard. From (2) and (3) and Leibniz theorem this series converge, but I don't know whether it will lead somewhere. Unfortunately rather Weierstrass M-test won't be helpful here.
 A: Essentially, such a series is alternating: for each $x\in D$, the series $\sum\limits_{n=1}^\infty{(-1)^{n+1}}f_n(x)$ is a convergent alternating series. From this, and 3) (note that 3) is saying that the sequence $(f_n)$ converges uniformly to $0$), it will follow that the series is uniformly Cauchy on $D$ and thus uniformly convergent on $D$.
Let $\alpha_n=\sup\limits_{x\in D}\,\{f_n(x)\}$.
Then for any positive integers $m$ and $n$ with $m\ge n$ and $x\in D$, using the fact that $\sum\limits_{n=1}^\infty{(-1)^{n+1}}f_n(x)$ is an alternating series of real numbers
$$\tag{1}
\Biggl|\,{(-1)^{n+1} } f_n(x)+{(-1)^{n+2} }f_{n+1}(x)+\cdots+{ (-1)^{m+1} }f_m(x)\,\Biggl|\ \le\  f_n(x)\le \alpha_n .
$$
The term on the right hand side of $(1)$ is independent of $x$ and can be made as small as desired,  since $\lim\limits_{n\rightarrow\infty }\alpha_n=0$.  From this, it follows that  the series $\sum\limits_{n=1}^\infty{(-1)^{n+1}}f_n(x)$ is uniformly Cauchy on $D$, and thus uniformly convergent on $D$.



Recall that a sequence $( g_n)$ of real-valued functions  is uniformly Cauchy on a set $A$ if for any $\epsilon>0$, there exists a positive integer $N$ so that whenever $n$ and $m$ are positive integers with $m,n\ge N$ we have $\bigl|g_n(x)-g_m(x)\bigr|<\epsilon$ for all $x\in A$.
An easily proven result is that a sequence $( g_n)$ of real-valued functions is uniformly convergent on $A$ if and only if it is uniformly Cauchy on $A$. A proof of this result can be found here.
This result phrased for a series of functions would read as follows: a series  $\sum\limits_{n=1}^\infty g_n$ of real valued functions converges uniformly on $A$ if and only if its sequence of partial sums is uniformly Cauchy on $A$.  
We say a series of functions is uniformly Cauchy if its sequence of partial sums is. 
Note that $\sum\limits_{n=1}^\infty g_n$ is uniformly Cauchy on $A$ if and only if for any $\epsilon>0$, there is a positive integer $N$ so that for $m\ge n\ge N$,
we have $\Bigl| \sum\limits_{k=n}^m g_k(x)\Bigr|<\epsilon$ for all $x\in A$.
