Show that 1n∑j=1∞(1−pj)n)→0 as n→∞

Step 1
From the inequality that OP observed, we get
${\displaystyle {(1-p_j)}^n\ge max\{1-n{p}_j,0\}}$
and hence
${\displaystyle 1-{(1-p_j)}^n\le min\{n{p}_j,1\}}$ From this, we get
${\displaystyle \frac{1}{n}\sum _{j=1}^{\mathrm{\infty }}(1-{(1-p_j)}^n)\le \sum _{j=1}^{\mathrm{\infty }}min\{p_j,\frac{1}{n}\}}$
Since the jth summand of the last sum is always bounded by ${\displaystyle p_j}$ and ${\displaystyle \sum _{j=1}^{\mathrm{\infty }}{p}_j}$ converges, by the dominated convergence theorem
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\sum _{j=1}^{\mathrm{\infty }}min\{p_j,\frac{1}{n}\}=\sum _{j=1}^{\mathrm{\infty }}=\sum _{j=1}^{\mathrm{\infty }}\underset{n\to \mathrm{\infty }}{lim}min\{p_j,\frac{1}{n}\}=\sum _{j=1}^{\mathrm{\infty }}min\{p_j,0\}=0}$
So by the squeezing theorem,
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}\frac{1}{n}\sum _{j=1}^n{(1-p_j)}^n)=0}$

Step 1
Applying the Stolz-Cesaro theorem, we have to compute
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}[\sum _{j=1}^{\mathrm{\infty }}(1-{(1-p_j)}^{n+1}-\sum _{j=1}^{\mathrm{\infty }}(1-{(1-p_j)}^n)]}$
${\displaystyle =\underset{n\to \mathrm{\infty }}{lim}\sum _{j=1}^{\mathrm{\infty }}{p}_j{(1-p_j)}^n}$
From the given sum it follows that ${\displaystyle p_j\le 1}$. Thus, ${\displaystyle 0\le p_j{(1-p_j)}^n\le p_j}$
and we know ${\displaystyle \sum _{j=1}^{\mathrm{\infty }}{p}_j}$ converges. This implies that we can switch the limit and the summation in the previous line.
${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{p}_j{(1-p_j)}^n=0}$
${\displaystyle \Rightarrow \underset{n\to \mathrm{\infty }}{lim}\sum _{j=1}^{\mathrm{\infty }}{p}_j{(1-p_j)}^n=\sum _{j=1}^{\mathrm{\infty }}0=0}$

Step 1 I think your bound suffices! In fact, we have that $\frac{1}{n}\sum _{j=k+1}^{\mathrm{\infty }}(1-p_j{)}^n)\le \frac{1}{n}\sum _{j=k+1}^{\mathrm{\infty }}n{p}_j=\sum _{j=k+1}^{\mathrm{\infty }}{p}_j$ This way, for each k we can bound the total sum as $\frac{1}{n}\sum _{j=1}^{\mathrm{\infty }}(1-(1-p_j{)}^n)=\frac{1}{n}\sum _{j=1}^k(1-p_j{)}^n)+\frac{1}{n}\sum _{j=k+1}^{\mathrm{\infty }}(1-(1-p_j{)}^n)$ $\le \frac{1}{n}|su{m}_{j=1}^{k}1+\sum _{j=k+1}^{\mathrm{\infty }}{p}_j$ $=\frac{k}{n}+\sum _{j=k+1}^{\mathrm{\infty }}{p}_j$ Now, the result follows by truncating. Fix $ϵ>0$. Take $k>0$ such that the sum $\sum _{j=k+1}^{\mathrm{\infty }}{p}_j$ is less than $\frac{ϵ}{2}$, and then take $n>\frac{2k}{ϵ}$. The computation from above shows that $\frac{1}{n}\sum _{j=1}^{\mathrm{\infty }}(1-p_j{)}^n)\le \frac{ϵ}{2}+\frac{ϵ}{2}=ϵ$ which proves the statement.