# A condition on convergence of an infinite product

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?

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

• 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. – gogurt Oct 8 '15 at 14:14
• @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. – user237392 Oct 8 '15 at 14:17