John Stewart

2022-01-15

Show that $\frac{1}{n}\sum _{j=1}^{\mathrm{\infty }}{\left(1-{p}_{j}\right)}^{n}\right)\to 0$ as $n\to \mathrm{\infty }$

Corgnatiui

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

intacte87

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

alenahelenash

Step 1 I think your bound suffices! In fact, we have that $\frac{1}{n}\sum _{j=k+1}^{\mathrm{\infty }}\left(1-{p}_{j}{\right)}^{n}\right)\le \frac{1}{n}\sum _{j=k+1}^{\mathrm{\infty }}n{p}_{j}=\sum _{j=k+1}^{\mathrm{\infty }}{p}_{j}$ This way, for each k we can bound the total sum as $\frac{1}{n}\sum _{j=1}^{\mathrm{\infty }}\left(1-\left(1-{p}_{j}{\right)}^{n}\right)=\frac{1}{n}\sum _{j=1}^{k}\left(1-{p}_{j}{\right)}^{n}\right)+\frac{1}{n}\sum _{j=k+1}^{\mathrm{\infty }}\left(1-\left(1-{p}_{j}{\right)}^{n}\right)$ $\le \frac{1}{n}|su{m}_{j=1}^{k}1+\sum _{j=k+1}^{\mathrm{\infty }}{p}_{j}$ $=\frac{k}{n}+\sum _{j=k+1}^{\mathrm{\infty }}{p}_{j}$ 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 $\frac{1}{n}\sum _{j=1}^{\mathrm{\infty }}\left(1-{p}_{j}{\right)}^{n}\right)\le \frac{ϵ}{2}+\frac{ϵ}{2}=ϵ$ which proves the statement.