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.

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