Convergence of a "shuffled" sequence

Question

Let $ \{\alpha_n \}_{n \in \mathbb{N}} \subseteq (0, 1) $ be a sequence. From it, define the following sequence $\{\tau_n\}_{n \in \mathbb{N}}$ by:

$$\tau_0 = 1, \tau_1 = 0, \tau_{n+2} = \alpha_n\tau_n + (1 - \alpha_n)\tau_{n+1}, \forall n \in \mathbb{N}.$$

In other words, $\tau_2=\alpha_0, \tau_3=\alpha_0(1 - \alpha_1)$ and: $$\tau_n = \alpha_0 + \sum_{j=1}^{n-2}(-1)^j \prod_{i=0}^{i=j} \alpha_i \quad \text { for all } n \geq 3.$$

Does $\{\tau_n\}$ converges?

My guess is yes. Here is what I did:

Let $m, n \in \mathbb{N}, m > n$. Then: $$|\tau_m - \tau_n| = |\alpha_0 + \sum_{j=1}^{m-2}(-1)^j \prod_{i=0}^{i=j} - \alpha_0 - \sum_{j=1}^{n-2}(-1)^j \prod_{i=0}^{i=j}| = |\sum_{j=n-1}^{m-2}(-1)^j \prod_{i=0}^{i=j}|. $$

I.e., $|\tau_m - \tau_n| = \alpha_{n - 1}...\alpha_{m - 2} \to 0$ for large enough $n, m$, since each $\alpha_i \in (0, 1)$.

By Cauchy's criterion, $\{\tau_n\}$ converges.

Is my argument correct?

I also tried this 'shuffling' argument and wrote the odds and even index terms $\{\tau_{2k+1}\}$ and $\{\tau_{2k}\}$, unsuccessfully.

Thanks.

Answer 1

Much of the argument is correct, but unfortunately there's a problem step (which there must be, because the statement to be proved isn't true...!):

It is not the case that $\alpha_{n-1}\cdots\alpha_{m-2}$ must go to $0$. Take for example $\alpha_k = 1-\frac1{(k+2)^2}$. Then the whole product $\prod_{k=1}^\infty \alpha_k$ converges to a positive constant.

If we were allowed $\alpha_k=1$ exactly, then it would be easy to see that the $\tau$ sequence doesn't converge—it would just oscillate back and forth between $1$ and $0$. Making $\alpha_k$ approach $1$ rapidly mimics this behaviour enough for the resulting sequence to still diverge. Here's a plot of the first $100$ terms of the $\tau$ sequence when $\alpha_k = 1-\frac1{(k+2)^2}$:

Answer 2

Note, $\tau_{n+2}$ is strictly between $t_n$ and $t_{n+1},$ and you get the width $$|\tau_{n+2}-\tau_{n+1}|=\alpha_n|\tau_{n+1}-\tau_n|.$$

So if $\prod_n \alpha_n \to 0,$ you definitely get a limit.

The closed intervals $I_n$ with endpoints $\tau_n$ and $\tau_{n+1}$ have the property $I_{n+1}\subseteq I_n,$ so the intersection of the $I_n$ is also a closed interval, $I=[a,b],$ possibly with $a=b.$

You'll get $a=b$ precisely when $\prod \alpha_n \to 0.$ More generally, $$b-a=\lim_{N\to\infty}\prod_{n=1}^N\alpha_n$$

You'll have, in general, $\liminf \tau_n = a$ and $\limsup \tau_n = b.$

There are plenty of examples of $\alpha_n$ with non-zero infinite product. For example, $\alpha_n=1-\frac 1{n^2}.$