# Proving $u_n(x)\le v_n(x), \sum\limits_{n=1}^{\infty}v_n(x)\to f$ uniformlly $\Longrightarrow\sum\limits_{n=1}^{\infty}u_n(x)\to g$ uniformly

Prove if $u_n(x)\le v_n(x)$ and $\sum\limits_{n=1}^{\infty}v_n(x)$ converges uniformly then also $\sum\limits_{n=1}^{\infty}u_n(x)$ converge uniformly

I thought solving it by using Weierstrass test. Since $\sum v_n(x)$ converges uniformly it can bounded by $\sum M_n$ which converge and then $\sum u_n(x)\le\sum v_n(x)\le \sum M_n$ which means $\sum u_n(x)$ converge uniformly. Is it correct?

• Sense your not operating at all with the variable $x$, it might be useful just to drop it all together. – Ethan Jul 7 '13 at 4:30

This is not correct. Not every uniformly convergent series is normally convergent (existence of such $M_n$ independent of $x$). For instance, consider $f_n(x)=\frac{1}{n}$ if $x=n$ and $f_n(x)=0$ elsewhere. This converges uniformly, but not normally (there does not exist a dominating convergent series $\sum M_n$), on $\mathbb{R}$. So Weierstrass M-test implies uniform convergence, but there are uniformly convergent series to which the Weierstrass M-test can not be applied.
I assume that $0\leq u_n(x)\leq v_n(x)$, otherwise, the result you mention is clearly false (take $u_n(x)=-1\leq 0=v_n(x)$).
To prove this result, it suffices to use the Cauchy criterion. A real-valued series of functions is uniformly convergent if and only if it is uniformly Cauchy, i.e. $$\forall \epsilon>0\quad\exists N\quad \forall n\geq N, k\geq 1\quad \Big|\sum_{j=1}^k v_{n+j}(x) \Big|\leq \epsilon.$$ Now just use that $0\leq u_n(x)\leq v_n(x)$ to prove that the uniform convergence of $\sum v_n$ implies that of $\sum u_n$.