Show that $\frac{1}{n}\sum_{j=1}^\infty \left( 1 - (1-p_j)^n\right) \to 0$ as $n \to \infty$

Given $p_j \geq 0$ for all $j \geq 1$ and $\sum_{j=1}^\infty p_j = 1$, I am asked to show that $$\frac{1}{n}\sum_{j=1}^\infty \left( 1 - (1-p_j)^n\right) \to 0 ~~ \textrm{as} ~~ n \to \infty.$$ Unfortunately, using only the fact that $(1-p_j)^n \geq 1 - np_j$ will not be enough, as this inequality only gives us $$\frac{1}{n}\sum_{j=1}^\infty \left( 1 - (1-p_j)^n\right) \leq \frac{1}{n}\sum_{j=1}^\infty np_j = 1.$$ Can anyone provide a hint towards the proof?

Remark: This problem is related to exercise 3.8 in this monograph.

I think your bound suffices! In fact, we have that $$ \frac{1}{n} \sum_{j=k+1}^\infty (1-(1-p_j)^n) \leq \frac{1}{n} \sum_{j=k+1}^\infty n p_j = \sum_{j=k+1}^\infty p_j. $$ This way, for each $k$ we can bound the total sum as \begin{align*} \frac{1}{n}\sum_{j=1}^\infty (1-(1-p_j)^n) &= \frac{1}{n} \sum_{j=1}^k (1-(1-p_j)^n) + \frac{1}{n}\sum_{j=k+1}^\infty (1-(1-p_j)^n) \\ &\leq \frac{1}{n}\sum_{j=1}^k 1+\sum_{j=k+1}^\infty p_j \\ &= \frac{k}{n} +\sum_{j=k+1}^\infty p_j. \end{align*} Now, the result follows by truncating. Fix $\varepsilon>0$. Take $k>0$ such that the sum $\sum_{j=k+1}^\infty p_j$ is less than $\varepsilon/2$, and then take $n>2k/\varepsilon$. The computation from above shows that $$ \frac{1}{n}\sum_{j=1}^\infty (1-(1-p_j)^n) \leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2}=\varepsilon, $$ which proves the statement.

From the inequality that OP observed, we get $(1-p_j)^n \geq \max\{1 - np_j, 0\}$ and hence

$$ 1 - (1-p_j)^n \leq \min\{np_j, 1\}, $$

From this, we get

$$ \frac{1}{n} \sum_{j=1}^{\infty} (1 - (1-p_j)^n) \leq \sum_{j=1}^{\infty} \min\{p_j, 1/n\}. $$

Since the $j$th summand of the last sum is always bounded by $p_j$ and $\sum_{j=1}^{\infty} p_j$ converges, by the dominated convergence theorem

$$ \lim_{n\to\infty} \sum_{j=1}^{\infty} \min\{p_j, 1/n\} = \sum_{j=1}^{\infty} \lim_{n\to\infty} \min\{p_j, 1/n\} = \sum_{j=1}^{\infty} \min\{p_j, 0\} = 0. $$

So by the squeezing theorem,

$$ \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} (1 - (1-p_j)^n) = 0. $$