Q:

Local uniform convergence of the terms of a sum is sufficient for convergence of the partial sum?

Consider  

a  mapping $u$ on $\mathbb{N} \mapsto \mathbb{R}$ with $k \mapsto u(k)\equiv u_k$, bounded in absolute value  by  $\bar{u}<\infty$.
a  mapping $a$ on $\mathbb{N} \mapsto (0,\infty)$ with $n\mapsto a(n)\equiv a_n$. 
a  mapping $b$ on $\mathbb{N} \mapsto \mathbb{R}$ with $n\mapsto b(n)\equiv b_n$. 

$\forall n \in \mathbb{N}$, $\forall t \in \mathbb{R}$, $\forall x\in \mathbb{R}$, consider the object
$$
n(1-G(a_n*(t+x)+b_n))
$$
where (not sure that all these details are needed)

$G:\mathbb{R}\rightarrow [0,1]$  
$G$ continuous and strictly monotone increasing on $\mathbb{R}$ 
$\lim_{x\rightarrow \infty}G(x)=1$ 
$\lim_{x\rightarrow -\infty}G(x)=0$ 
$G$ is such that 
$$
\exists \text{ }A:\mathbb{R}\rightarrow (0,\infty) \text{ s.t. } \lim_{s\rightarrow \infty }\frac{1-G(s+v A(s))}{1-G(s)} = e^{-v}\text{ }\forall v \in \mathbb{R}
$$
Moreover, $A(s)\equiv \frac{1-G(s)}{g(s)}$ where $G(s)\equiv \int_{-\infty}^s g(e) de$ $\forall s \in \mathbb{R}$. 
$b_n\equiv G^{-1}(1-\frac{1}{n})$, so that $\lim_{n\rightarrow \infty}b_{n}=\infty$ (which implies $\lim_{n\rightarrow \infty}G(b_{n})=1$)
$a_n\equiv A(b_n)$. 
Under these conditions it can be shown that $\lim_{n\rightarrow \infty} G(a_n*(t+x)+b_n)=1$ $\forall x\in \mathbb{R}$ $\forall t \in \mathbb{R}$.

Assume that 
$$
\lim_{n\rightarrow \infty} n(1-G(a_n*(t+x)+b_n))= \exp(-t-x) \text{ }\text{ $\forall t \in \mathbb{R}$ $\forall x\in \mathbb{R}$}
$$
I want to show that  $\forall t \in \mathbb{R}$
$$
\lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^{2n}[n(1-G(a_n*(t+u_k)+b_n))- \exp(-t-u_k)]=0
$$

The book gives two hints:
1) We can see that $\forall t \in \mathbb{R}$ the function $x\mapsto n(1-G(a_n*(t+x)+b_n))$ converges locally uniform to $\exp(-t-x)$ as $n\rightarrow \infty$ (application of result 0.1 in Resnick's book).
2) Notice that $\forall t \in \mathbb{R}$
$$
 \frac{1}{n}\sum_{k=1}^{2n}\exp(-t-u_k)]\leq 1+\exp(\bar{u})<\infty \text{ }\text{ }\forall n\in \mathbb{N} 
$$

Which result (with 1) and 2) as sufficient conditions) the book wants me to consider? Could you give some further references or a proof if short? 

A:

You've essentially pointed out all of the needed. The fact that, for any $v\in\mathbb{R}$,
$$\lim_{n\to\infty}n\big(1-G(a_n v+b_n)\big)=\lim_{n\to\infty}\frac{1-G(b_n+vA(b_n))}{1-G(b_n)}=e^{-v},$$
follows from the conditions on $G$ (it doesn't need to be "assumed" for $v=t+x$).
The result referenced by you is that this convergence is locally uniform, meaning that if $$M_n(v_1,v_2)=\sup_{v\in[v_1,v_2]}\Big|n\big(1-G(a_n v+b_n)\big)-e^{-v}\Big|,$$
then $\displaystyle\lim_{n\to\infty}M_n(v_1,v_2)=0$, and this is perfectly enough since, obviously,
$$\left|\frac{1}{n}\sum_{k=1}^{2n}\Big(n\big(1-G(a_n(t+u_k)+b_n)\big)-e^{-t-u_k}\Big)\right|\leqslant 2M_n(t-\bar{u},t+\bar{u}).$$

