6
$\begingroup$

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.

$\endgroup$
9
  • $\begingroup$ @KentaS I don't think so by a quick try... If you could do it can you show me the details...? For someone who voted "close", may I know the reason? $\endgroup$
    – Fei Cao
    Dec 23, 2021 at 3:56
  • $\begingroup$ I did not vote to close, but the reason given was lack of context. $\endgroup$
    – robjohn
    Dec 23, 2021 at 8:33
  • $\begingroup$ If you provide some context, other answers will most likely be posted. Furthermore, it should prevent more close votes and the probable delete votes following. I think this is a good question, it just needs to be expanded to meet the site requirements. Where did you get the problem? Did this come up in a course/book; if so which course/book? What tools are being discussed in the course/book? etc. $\endgroup$
    – robjohn
    Dec 23, 2021 at 16:54
  • $\begingroup$ Even if, in your question, you just showed that $(1-p_j)^n \ge1-np_j$, as is, only seems to show that $\lim\limits_{n\to\infty}\frac1n\sum\limits_{j=1}^\infty\left(1-(1-p_j)^n\right)\le\sum\limits_{j=1}^\infty p_j=1$, that would add more context. $\endgroup$
    – robjohn
    Dec 24, 2021 at 13:40
  • 1
    $\begingroup$ @robjohn I have taken care of your suggestions and the original post has been modified~ Merry Christmas! $\endgroup$
    – Fei Cao
    Dec 25, 2021 at 1:12

5 Answers 5

11
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ (+1) Nice solution! I posted another solution that is inspired by your approach. $\endgroup$ Dec 23, 2021 at 6:31
  • $\begingroup$ This solution is just so NICE! $\endgroup$
    – Fei Cao
    Dec 23, 2021 at 21:59
5
$\begingroup$

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,

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

$\endgroup$
1
$\begingroup$

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

$\endgroup$
1
$\begingroup$

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 $$ \lim_{n\to\infty}\frac1n\sum_{j=1}^\infty\left(1-(1-p_j)^n\right)=0\tag5 $$

$\endgroup$
1
$\begingroup$

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

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .