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.
 A: 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,
$$ \lim_{n\to\infty} \frac{1}{n} \sum_{j=1}^{n} (1 - (1-p_j)^n) = 0. $$
A: 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.
A: Applying the Stolz-Cesaro theorem, we have to compute
$$\lim_{n\to\infty}\left[\sum_{j=1}^{\infty}\left(1-(1-p_j)^{n+1}\right)-\sum_{j=1}^{\infty}\left(1-(1-p_j)^{n}\right)\right]
\\=\lim_{n\to\infty}\sum_{j=1}^{\infty}p_j(1-p_j)^n $$
From the given sum it follows that $p_j\le 1$.
Thus, $0\le p_j(1-p_j)^n\le p_j$, and we know $\sum_{j=1}^{\infty}p_j$ converges. This implies that we can switch the limit and the summation in the previous line.
$$\lim_{n\to\infty}p_j(1-p_j)^n=0 \\
\Rightarrow \lim_{n\to\infty}\sum_{j=1}^{\infty}p_j(1-p_j)^n=\sum_{j=1}^{\infty}0=0 $$
A: Here is another approach to splitting the sum at a strategic point.

Dominated Convergence says that
$$
\lim_{k\to\infty}\sum_{j=1}^\infty(1-p_j)^kp_j=0\tag1
$$
Thus, for any $\epsilon\gt0$, there is a $k_\epsilon$ so that for all $k\ge k_\epsilon$
$$
\sum_{j=1}^\infty(1-p_j)^kp_j\le\epsilon\tag2
$$
Therefore, for $n\ge k_\epsilon$,
$$
\begin{align}
\frac1n\sum_{j=1}^\infty\left(1-(1-p_j)^n\right)
&=\frac1n\sum_{j=1}^\infty\frac{1-(1-p_j)^n}{1-(1-p_j)}p_j\tag{3a}\\
&=\frac1n\sum_{j=1}^\infty\sum_{k=0}^{n-1}(1-p_j)^kp_j\tag{3b}\\
&=\frac1n\sum_{k=0}^{n-1}\sum_{j=1}^\infty(1-p_j)^kp_j\tag{3c}\\
&=\frac1n\sum_{k=0}^{k_\epsilon-1}\sum_{j=1}^\infty(1-p_j)^kp_j+\frac1n\sum_{k=k_\epsilon}^{n-1}\sum_{j=1}^\infty(1-p_j)^kp_j\tag{3d}\\
&\le\frac1n\ \underbrace{\sum_{k=0}^{k_\epsilon-1}\sum_{j=1}^\infty(1-p_j)^kp_j}_\text{constant in $n$}+\underbrace{\ \frac{n-k_\epsilon}{n}\ \vphantom{\sum_j^1}}_{\le1}\ \epsilon\tag{3e}
\end{align}
$$
Explanation:
$\text{(3a)}$: $1-(1-p_j)=p_j$
$\text{(3b)}$: $\frac{1-x^n}{1-x}=\sum\limits_{k=0}^{n-1}x^k$
$\text{(3c)}$: swap order of summation
$\text{(3d)}$: split the outer sum at $k_\epsilon$
$\text{(3e)}$: apply $(2)$
Taking the limit of $(3)$,
$$
\lim_{n\to\infty}\frac1n\sum_{j=1}^\infty\left(1-(1-p_j)^n\right)
\le\epsilon\tag4
$$
Since $\epsilon\gt0$ was arbitrary, $(4)$ means
$$
\lim_{n\to\infty}\frac1n\sum_{j=1}^\infty\left(1-(1-p_j)^n\right)=0\tag5
$$
A: I came to know this $\epsilon$-$N$ argument from the elementary proof of Tannery theorem [1].
(I used Tannery theorem to prove that
$\lim_{n \to \infty} \int^1_0 f(x) |\sin(n \pi x)|\,\mathrm{d}x = \frac{2}{\pi} \int^1_0 f(x)\,\mathrm{d}x$ Proving $\lim_{n \to \infty} \int^1_0 f(x) |\sin(n \pi x)|\,dx = \frac{2}{\pi} \int^1_0 f(x)\,dx$ .)
Denote $m = \lfloor \sqrt{n}\rfloor$ where $\lfloor \cdot \rfloor$ is the floor function. We have
\begin{align*}
 &1 - (1 - p_j)^n\\
 =\,& p_j\Big(1 + (1 - p_j) + (1 - p_j)^2 + \cdots + (1 - p_j)^{n - 1}\Big)\\
 \le\,& p_j \Big(1 + (1 - p_j) + (1 - p_j)^2 + \cdots + (1 - p_j)^{m - 1} + n (1 - p_j)^m\Big)\\
 \le\,& p_j \Big(m +  n(1 - p_j)^m\Big).
\end{align*}
Thus,
$$\frac{1}{n}[1 - (1 - p_j)^n] \le \frac{m}{n} p_j + p_j(1 - p_j)^m.$$
It suffices to prove that
$$\lim_{m\to \infty} \sum_{j=1}^\infty p_j(1 - p_j)^m = 0.$$
$\epsilon$-$N$ argument:
For any given $\epsilon > 0$, there is
an $N$ such that
$\sum_{j > N} p_j(1 - p_j)^m
\le \sum_{j > N} p_j < \frac{\epsilon}{2}$.
Clearly, for each $j$, there is an $M_j$ such that
$p_j(1 - p_j)^m < \frac{\epsilon}{2N}$
for all $m > M_j$.
Let $M = \max(M_1, M_2, \cdots, M_N)$.
We have
$$\sum_{j=1}^\infty p_j(1 - p_j)^m
= \sum_{j\le N} p_j(1 - p_j)^m
+ \sum_{j > N} p_j(1 - p_j)^m
< N \cdot \frac{\epsilon}{2N} + \frac{\epsilon}{2} = \epsilon$$
for all $m > M$.
Reference:
[1] Josef Hofbauer, "A simple proof of 1 + 1/22 + 1/32 + ... = PI2/6 and related identities",
American Mathematical Monthly 109 (February 2002), 196-200.
https://homepage.univie.ac.at/josef.hofbauer/piq6.htm
