John Stewart

2022-01-15

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

Corgnatiui

Beginner2022-01-16Added 35 answers

Step 1

From the inequality that OP observed, we get

$(1-{p}_{j})}^{n}\ge max\{1-n{p}_{j},0\$

and hence

$1-{(1-{p}_{j})}^{n}\le min\{n{p}_{j},1\}$
From this, we get

$\frac{1}{n}\sum _{j=1}^{\mathrm{\infty}}(1-{(1-{p}_{j})}^{n})\le \sum _{j=1}^{\mathrm{\infty}}min\{{p}_{j},\frac{1}{n}\}$

Since the jth summand of the last sum is always bounded by$p}_{j$ and $\sum _{j=1}^{\mathrm{\infty}}{p}_{j}$ converges, by the dominated convergence theorem

$\underset{n\to \mathrm{\infty}}{lim}\sum _{j=1}^{\mathrm{\infty}}min\{{p}_{j},\frac{1}{n}\}=\sum _{j=1}^{\mathrm{\infty}}=\sum _{j=1}^{\mathrm{\infty}}\underset{n\to \mathrm{\infty}}{lim}min\{{p}_{j},\frac{1}{n}\}=\sum _{j=1}^{\mathrm{\infty}}min\{{p}_{j},0\}=0$

So by the squeezing theorem,

$\underset{n\to \mathrm{\infty}}{lim}\frac{1}{n}\sum _{j=1}^{n}{(1-{p}_{j})}^{n})=0$

From the inequality that OP observed, we get

and hence

Since the jth summand of the last sum is always bounded by

So by the squeezing theorem,

intacte87

Beginner2022-01-17Added 42 answers

Step 1

Applying the Stolz-Cesaro theorem, we have to compute

$\underset{n\to \mathrm{\infty}}{lim}[\sum _{j=1}^{\mathrm{\infty}}(1-{(1-{p}_{j})}^{n+1}-\sum _{j=1}^{\mathrm{\infty}}(1-{(1-{p}_{j})}^{n})]$

$=\underset{n\to \mathrm{\infty}}{lim}\sum _{j=1}^{\mathrm{\infty}}{p}_{j}{(1-{p}_{j})}^{n}$

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$\sum _{j=1}^{\mathrm{\infty}}{p}_{j}$ converges. This implies that we can switch the limit and the summation in the previous line.

$\underset{n\to \mathrm{\infty}}{lim}{p}_{j}{(1-{p}_{j})}^{n}=0$

$\Rightarrow \underset{n\to \mathrm{\infty}}{lim}\sum _{j=1}^{\mathrm{\infty}}{p}_{j}{(1-{p}_{j})}^{n}=\sum _{j=1}^{\mathrm{\infty}}0=0$

Applying the Stolz-Cesaro theorem, we have to compute

From the given sum it follows that

and we know

alenahelenash

Expert2022-01-24Added 556 answers

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 $\u03f5>0$ . Take $k>0$ such that the sum $\sum _{j=k+1}^{\mathrm{\infty}}{p}_{j}$ is less than $\frac{\u03f5}{2}$ , and then take $n>\frac{2k}{\u03f5}$ . The computation from above shows that
$\frac{1}{n}\sum _{j=1}^{\mathrm{\infty}}(1-{p}_{j}{)}^{n})\le \frac{\u03f5}{2}+\frac{\u03f5}{2}=\u03f5$
which proves the statement.

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