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My question is about the convergence of sums of two-dimensional arrays of reals. What I mean by a two-dimensional array is a mapping on $\mathbb{N}\times\mathbb{N} \mapsto \mathbb{R}$. Let $b_{n,k}$ be such a mapping, i.e. $(n,k) \mapsto b_{n,k}$.

Let $$ a_n\equiv \sum_{k=1}^{2n} \log(G(b_{n,k})) $$ and $$ \tilde{a}_n \equiv \sum_{k=1}^{2n} (1-G(b_{n,k})) $$ Assume also that

  • $G:\mathbb{R}\rightarrow [0,1]$ ,

  • $G$ continuous and strictly monotone increasing on $\mathbb{R}$.

  • $\lim_{x\rightarrow \infty}G(x)=1$

  • $\lim_{x\rightarrow -\infty}G(x)=0$

  • $\lim_{n\rightarrow \infty}b_{n,k}=\infty$ $\forall k$ (which implies $\lim_{n\rightarrow \infty}G(b_{n,k})=1$ $\forall k$)

I want to show $$ \lim_{n\rightarrow \infty}(a_n+\tilde{a}_n)=0 $$ and your hint would be extremely appreciated

This is what I have tried to do and where I am stuck:

Step 1: $$ \begin{cases} \lim_{n\rightarrow \infty} \log(G(b_{n,k}))=0 & \forall k\in 1,...,2n\\ \lim_{n\rightarrow \infty} (1-G(b_{n,k}))=0 & \forall k\in 1,...,2n\\ \end{cases} $$

Step 2 (wrong as explained here): from Step 1 it follows $$ \begin{cases} \lim_{n\rightarrow \infty}\sum_{k=1}^{2n} \log(G(b_{n,k}))=0 \\ \lim_{n\rightarrow \infty} \sum_{k=1}^{2n} (1-G(b_{n,k}))=0 \\ \end{cases} $$ which implies $$ \lim_{n\rightarrow \infty}\Big[\sum_{k=1}^{2n}\log(G(b_{n,k}))+\sum_{k=1}^{2n} (1-G(b_{n,k}))\Big]=0 $$

Also this question on the possibility of exchanging limit and infinite summation may be related.

Star
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  • If $G$ is actually $\to (0,1)$, we can take $c_n$ with $\lim_{n\to\infty}c_n=\infty$ and $G(c_n)<1-1/\sqrt{n}$ (say). And then your statement fails (with $b_{n,k}=c_n$ independent on $k$). You're really trying to prove something more specific. – metamorphy Apr 23 '18 at 16:32
  • @metamorphy Thanks for your help. Is your example compatible with the assumptions I have added? Why? – Star Apr 23 '18 at 16:55
  • Also: the hint that I received from the professor is: for $G\rightarrow 1$ we have that $-\log(G)\approx 1-G$ – Star Apr 23 '18 at 17:02
  • I've posted it as an answer. – metamorphy Apr 23 '18 at 17:21

1 Answers1

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Take $c_n=\sup\{c\in\mathbb{R}\ :\ G(c) \leqslant 1-1/\sqrt{n}\}$. Then $c_n$ is monotonically nondecreasing (clearly); $\lim_{n\to\infty}c_n = \infty$ (otherwise, if this limit is $c\in\mathbb{R}$, by continuity we would have $G(c)=1$); finally, $G(c_n) \leqslant 1-1/\sqrt{n}$ (again by continuity).

Now take $b_{n,k}=c_n$ independent on $k$. As $1-x+\log{x}<-(1-x)^2/2$ for $x\in(0,1)$, we get $a_n+\tilde{a}_n<-1$ for all $n$. The statement fails.

metamorphy
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  • Thanks. Could you clarify how you go from $1-x+\log(x)<...$ to $a_n+\tilde{a}_n<-1$? – Star Apr 23 '18 at 17:38
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    $a_n+\tilde{a}_n$ is the sum of $2n$ identical terms equal to $1-x+\log{x}$ with $x=1-1/\sqrt{n}$, which is $<-1/(2n)$. – metamorphy Apr 23 '18 at 17:42
  • why with $x=1-\frac{1}{\sqrt{n}}$? Should it be with $x\leq 1-\frac{1}{\sqrt{n}}$? – Star Apr 23 '18 at 17:46
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    Ok, sorry, take $\leqslant$ instead of $=$ ;) The inequality still holds. – metamorphy Apr 23 '18 at 17:47
  • Sorry, I am still confused. For $n$ large we have that $\log(1-\frac{1}{\sqrt{n}})=-[ 1-(1-\frac{1}{\sqrt{n}})]$ (which is the hint of my professor). How is that related to your answer? Also, does something change if our objective is to claim $a_n=-\tilde{a}_n+o(1)$ where $o(1)$ is a quantity going to $0$ as $n\rightarrow \infty$? – Star Apr 23 '18 at 19:27
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    Your professor is right in the sense that $-\log(1-x)=\displaystyle\sum_{k=1}^\infty\frac{x^k}{k}$ for $x\in(0,1)$. So $-x-\log(1-x)$ is at least $\displaystyle\frac{x^2}{2}$. This is "not small enough". We took $x_n=n^{-1/2}$ and got $|a_n+\tilde{a}_n|>1$. I could take $x_n=n^{-\epsilon}$ and get $|a_n+\tilde{a}_n|>n^{1-2\epsilon}$. This is not only not $o(1)$ - this may grow unbounded. – metamorphy Apr 23 '18 at 19:44
  • Ok, if I take $x_n\equiv n^{-1}$, then it works? – Star Apr 23 '18 at 19:54
  • Let us continue this discussion in chat. – metamorphy Apr 23 '18 at 19:55
  • Please give a look to my related question, it is very similar but I am imposing a specific structure to $b_{n,k}$ https://math.stackexchange.com/questions/2750587/convergence-of-the-sum-of-two-infinite-sums – Star Apr 25 '18 at 07:48
  • @CGT I should have defined $c_n$ this way for $n>1$ (as we analyse $n\to\infty$, this really doesn't matter). In this setting, the set in question (omit the $\sup$) is nonempty due to conditions on $G$ (I don't impose anything in addition to what you did - and I don't understand what do you mean saying that a function is growing slower than a sequence...). As of the related question - a bit later... I'm kinda short of time :( – metamorphy Apr 25 '18 at 17:21