I have the following recursion of sequence $\{A_i\}_{i=1}^\infty$ and $\{B_i\}_{i=0}^\infty$


1 $~~~~~~~~~~~~~~A_1=q\sigma$ $~~~~~~~~~~~~~~~~~~~~~~~~~~~~~B_1=B_0-A_1B_0/\sigma$

2 $~~~~~~~~~~~~~~A_2=A_1(1-\sigma)$ $~~~~~~~~~~~~~~~~B_2=B_1-A_2B_1/\sigma$

$n$ $~~~~~~~~~~~~~A_n=A_{n-1}(1-\sigma)$ $~~~~~~~~~~~~B_n=B_{n-1}-A_nB_{n-1}/\sigma$

where $B_0=\sigma\in(0,1)$ and $q\in(0,1)$.

So, it is clearly that $A_n=q\sigma(1-\sigma)^{n-1}$. But $B_n$ is difficult to get a simple form.

Then I want to compute the following series \begin{align*} \sum_{i=1}^{\infty}(1-\sigma)^{i-1}B_{i-1}&=\frac{1}{q}\sum_{i=1}^{\infty}A_iB_{i-1}/\sigma\\ &=\frac{1}{q}(B_0-B_1+B_1-B_2+B_2-B_3+....)\\ &=\frac{1}{q}B_0=\sigma/q \end{align*}

The last step is questionable, as it is an alternating series. One can show that $B_n$ is decreasing and is bounded by $\sigma$. So if you replace $B_n$ by $\sigma$, then this whole series is 1. So the originaly series is bounded by 1. But $B_n$ does not go to zero as $n$ increases. So it fails one of the alternating series convergence test. I know that it is a sufficient condition.

Then I am stuck at this point. How to solve this series? Is this even convergent?

  • 1
    $\begingroup$ If $\{B_n\}$ is decreasing, then $L=\lim_{n\to\infty} B_n$ exists, and then $\lim_{n\to\infty} \frac1q(B_0-B_1+B_1-B_2+B_2-B_3+\cdots+B_{n-1}-B_n) = \lim_{n\to\infty} \frac1q(B_0-B_n) = \frac1q(B_0-L)$. So it suffices to calculate $L$. $\endgroup$ – Greg Martin Jul 2 '18 at 7:25
  • $\begingroup$ @GregMartin Thanks! I see. But unfortunately, this limit of $B_n$ has no closed-form solution. But your method is great, I didn't think about it that way originally. Thanks! $\endgroup$ – ftxx Jul 2 '18 at 7:47
  • $\begingroup$ We certainly have $B_n = B_0 \prod_{j=0}^{n-1} \big(1-q(1-\sigma)^j \big)$ from its recursive definition; and $\prod_{j=0}^\infty \big(1-q(1-\sigma)^j \big)$ converges (because $\sum_{j=0}^\infty q(1-\sigma)^j$ converges); therefore $L = B_0 \prod_{j=0}^\infty \big(1-q(1-\sigma)^j \big)$. $\endgroup$ – Greg Martin Jul 2 '18 at 17:09

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