Say I have a sequence of numbers $(a_n)_{n \geq 1}$ such that $a_n \in (0,1)$ for all $n$ and $a_n \downarrow 0$. Suppose I know that $$ \prod_{k=1}^n a_k^{\beta^k} \to c > 0 $$ Where $\beta$ is some positive constant less than 1. Then does this give me any extra information about the convergence of: $$ \prod_{k=1}^n a_k^{\beta^{(n-k)}}$$ My sense is that it must tell me something, I'm just not sure what. Because even without the power $\beta^k$, the product $\prod^n a_k \to 0$ since $a_n \downarrow 0$. But now the first display says that deflating the $a_k$'s by $\beta^k$ we get convergence to a positive number. So in essence this limits the rate that $a_k \downarrow 0$ somehow. Is there any intuition here?

  • $\begingroup$ Did you try to use $\log$ on the product? $\endgroup$ – GBQT Oct 7 '15 at 17:31
  • $\begingroup$ @GBQT Yes, but it didn't result in anything immediately obvious. I'll take another look! $\endgroup$ – gogurt Oct 7 '15 at 18:57

Let's restate your first product in terms of logarithms:

$$a_k \downarrow 0: \lim_{n\to \infty}\sum_{k=1}^n \beta^k\log a_k \to \log c \in \mathbb{R}$$

Intuitively, $\beta^k \downarrow 0$ so that the diverging $\log a_k$ values are kept in check; therefore $0\leq \beta<1$.

Now, lets look at the modified version in your post:

$$\lim_{n\to \infty}\sum_{k=1}^n \beta^{n-k}\log a_k$$

What is different here is that: $\forall n, \lim\limits_{k\to n}\beta^{n-k} = 1$ so $\beta^{n-k}$ is a monotonic increasing sequence. However, we have left $\log a_k$ untouched, so:

$$\lim_{n\to \infty} \left[\lim_{k\to n} \beta^{n-k}\log a_k\right] = \lim_{n\to \infty} \log a_n = -\infty$$

Therefore, the last term in each finite sequence defined by a particular $n$ becomes negative without bound as a function of $n$ (due to $a_k \to 0$).

Formally we can use the Divergence Test for series:

$$\lim_{n\to \infty}a_n \neq 0 \implies \sum a_n \;\textrm{diverges} $$

Therefore, given that $\beta$ allows the first sequence to converge to a positive number, your second series will approach $-\infty$ if $\beta \neq 0$. Divergence to $-\infty$ in the logs implies that your product will converge to $0$ for any $0<\beta <1$ that satisfies your first product.

Convergence rates

You are correct that the convergence of the first product places limits on how fast $a_k \to 0$, since it cannot outstrip the rate that $\beta^k$ deflates the negative values in log space. To be more precise:

$$\beta^k\log a_k = O\left(\frac{-1}{k^p}\right), p>1$$

Therefore, for some $p>1, M>0$:

$$\log a_k = O\left(\frac{-1}{\beta^{k}k^p}\right) \implies a_k = O\left(\exp\left(\frac{-M}{\beta^{k}k^p}\right)\right)$$

Note that we cannot omit the constant in the implication, since it is no longer a multiplicative constant, but an exponential constant.

  • $\begingroup$ Thanks @Bey! The only thing is that it seems like your argument in the first section that the second product $\to 0$ doesn't actually depend on the convergence of the first product to a positive limit, right? The convergence of the first product really only plays a role in the rates, it seems. $\endgroup$ – gogurt Oct 8 '15 at 14:14
  • $\begingroup$ @gogurt you are correct, the fact that $\beta^{n-k}\log a_k$ decreases without bound in the $n^{th}$ term is sufficient. I was using the convergence of the first product series to bound $\beta$ which allowed me to not worry about the general case. $\endgroup$ – user237392 Oct 8 '15 at 14:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.