# Durret 2.2.1 and the statement “every convergent sequence is bounded”

Actually I asked this question on Math Stack Exchange (here) but it did not generate much interest. Since it is a question from probability theory, perhaps the crowd on cross validated will be more interested...(maybe this is incorrect etiquette to post the same question twice, but I am interested in the answer...)

The following is from Durrett's Probability Theory and Examples question 2.2.1:

Let $X_{1}, X_{2}, \ldots$ be uncorrelated with $\mathbb{E} X_{i} = \mu_{i}$ and $\frac{Var X_{i}}{i} \to 0$ as $i\to \infty$. Let $S_{n} = X_{1}+X_{2} + \ldots + X_{n}$ and $\nu_{n} = \mathbb{E} S_{n}/n$. Then as $n\to \infty$, $S_{n}/n - \nu_{n} \to 0$ in $L^{2}$ and in probability.

It can be shown that \begin{align*} \mathbb{E}\left[\left(\frac{S_{n}}{n} -\nu_{n} \right)^{2} \right] = \frac{1}{n^{2}} \sum_{i=1}^{n} Var(X_{i}) \end{align*} Now I want to show that $\frac{1}{n^{2}} \sum_{i=1}^{n} Var(X_{i}) \to 0$. My approach to this is for any fixed $\varepsilon>0$ I can find $N$ such that for all $i\geq N$ I have $Var(X_{i})/i < \varepsilon$. Now, if I knew $Var (X_{i})/i \leq M<\infty$ for $i=1, \ldots N$ then I can have \begin{align*} \frac{1}{n^{2}} \sum_{i=1}^{n} Var(X_{i}) &\leq \frac{1}{n} \sum_{i=1}^{n} \frac{Var(X_{i})}{i}\\ &=\frac{1}{n} \sum_{i=1}^{N} \frac{Var(X_{i})}{i} + \frac{1}{n} \sum_{i=N+1}^{n} \frac{Var(X_{i})}{i}\\ &\leq \frac{NM}{n} + \frac{(n-N-1)\varepsilon}{n} \end{align*} so that the result follows by taking $n\to \infty$ and noting that $\varepsilon$ is arbitrary. The issue is I don't know that there exists an $M$ such that $Var (X_{i})/i \leq M<\infty$ for $i=1, \ldots N$...

Presumably we can have $Var (X_{i})/i =\infty$? Is there anyway to prove that this is not the case? Or is my approach to the proof incorrect? Any help is appreciated.

• I understand that any convergent sequence of real numbers is bounded. But the proof that any convergent sequence of real numbers is bounded relies on the fact that the first $N$ members of the sequence are finite (which is fine since they are real numbers). My problem is that I do not know if $\mathbb{E} X_{i}^{2} <\infty$ so I do not know if $Var(X_{i})/i < \infty$. I feel there is nothing restricting $Var(X_{i})/i$ to be less than $\infty$.
• It's true that $\text{Var}(X_i)$ could be infinite, but only for finitely-many $i$ since $\text{Var}(X_i) / i \to 0$. You might try breaking up the sum into two pieces, one that contains all the terms with infinite variance and another that contains all the rest. The first sum has to go to zero when divided by $n$, and for the remaining terms you might try applying reasoning like the above. – dsaxton Nov 10 '16 at 15:57
• @dsaxton Why would the first sum (terms with infinite variance) have to go to zero when divided by $n$? – möbius Nov 10 '16 at 16:14
• Your last comment is a little puzzling, because no proof of convergence requires any property whatsoever to hold for some finite initial segment of the sequence. That is because convergence is a property that is logically independent of any such finite segment. If you want to work with extended reals rigorously, then (a) you must explicitly rule out convergence to $\infty$ and (b) remove any finite initial segment that might contain any value of $\infty$. You will obtain the desired result: any sequence of extended reals converging to a finite value eventually is bounded. – whuber Nov 10 '16 at 16:29
• @dsaxton How do I know the initial sum is finite? I agree that if $X_{i}$ is Borel measurable than it must be finite, but to take the limit as $n\to \infty$ inside the expectation operator I need to use either the dominated or monotonic convergence theorem. I cannot see an easy way to do this... let me know your thoughts – möbius Nov 10 '16 at 16:37
If $\operatorname{Var}X_k=\infty$ for any $k$, then the assertion isn't true since $\mathbb{E}\left[\left(\frac{S_{n}}{n} -\nu_{n} \right)^{2} \right]$, which equals $\frac{1}{n^{2}} \sum_{i=1}^{n} \operatorname{Var}(X_{i})$, will then be infinite for all sufficiently large $n$, and therefore we can't have convergence in $L^2$. So it is required that each $X_i$ has finite variance, and your proof goes through.