Find out the answer to "Show that 1n∑j=1∞(1−pj)n)→0 as n→∞"

John Stewart

Answered question

2022-01-15

Show that

\frac{1}{n} \sum_{j = 1}^{\infty} {(1 - p_{j})}^{n}) \to 0

n \to \infty

Corgnatiui

Beginner2022-01-16Added 35 answers

Step 1
From the inequality that OP observed, we get

{(1 - p_{j})}^{n} \geq max {1 - n p_{j}, 0}

and hence

1 - {(1 - p_{j})}^{n} \leq min {n p_{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}, \frac{1}{n}}

Since the jth 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}, \frac{1}{n}} = \sum_{j = 1}^{\infty} = \sum_{j = 1}^{\infty} lim_{n \to \infty} min {p_{j}, \frac{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 - p_{j})}^{n}) = 0

intacte87

Beginner2022-01-17Added 42 answers

Step 1
Applying the Stolz-Cesaro theorem, we have to compute

lim_{n \to \infty} [\sum_{j = 1}^{\infty} (1 - {(1 - p_{j})}^{n + 1} - \sum_{j = 1}^{\infty} (1 - {(1 - p_{j})}^{n})]

= lim_{n \to \infty} \sum_{j = 1}^{\infty} p_{j} {(1 - p_{j})}^{n}

From the given sum it follows that

p_{j} \leq 1

. Thus,

0 \leq p_{j} {(1 - p_{j})}^{n} \leq 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

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}^{\infty} (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

\frac{1}{n} \sum_{j = 1}^{\infty} (1 - (1 - p_{j})^{n}) = \frac{1}{n} \sum_{j = 1}^{k} (1 - p_{j})^{n}) + \frac{1}{n} \sum_{j = k + 1}^{\infty} (1 - (1 - p_{j})^{n})

\leq \frac{1}{n} | s u m_{j = 1}^{k} 1 + \sum_{j = k + 1}^{\infty} p_{j}

= \frac{k}{n} + \sum_{j = k + 1}^{\infty} p_{j}

Now, the result follows by truncating. Fix

ϵ > 0

. Take

k > 0

such that the sum

\sum_{j = k + 1}^{\infty} p_{j}

is less than

\frac{ϵ}{2}

, and then take

n > \frac{2 k}{ϵ}

. The computation from above shows that

\frac{1}{n} \sum_{j = 1}^{\infty} (1 - p_{j})^{n}) \leq \frac{ϵ}{2} + \frac{ϵ}{2} = ϵ

which proves the statement.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?