Consider the following recursive equation:

$$a(K) = K \frac{a(K-1) + b}{a(K-1) + b + S},K=1, 2, \cdots $$ where $b \in \{1, 2, \cdots \}$ and $S>0$ are both constants.

Is there any way to solve this recursive equation any give a closed-form $a(K)$? Or some tight upper and lower bound? (for example, one possible way to bound (a(K)) is using a(K +1) > a(K), then $a(K) \leq K \frac{a(K) + b}{a(K) + b + S}$).

I can simplify the equation by using $a(K) + b(K) = K$,

$$b(K) = \frac{KS}{b + S + K-1 - b(K-1)}$$

  • 1
    $\begingroup$ Assuming that $a(0)\ge 0$, we have that the sequence $a(k)/k$ consists of non-negative values, is strictly descending for $k\ge 2b$ and converges to $0$. First you prove that $a(k)/k\ge 0$ for all $k$ with induction, then you conclude that $a(k)/k\le 1-\frac{S}{b+S}$ for all $k\ge 1$. Then you prove it's descending, apply the MCT to conclude $c:=\lim_{k\to\infty}a(k)/k=\inf_{k\ge 1}a(k)/k$ exists. Assume $c>0$, conclude that $c\ge 1$ and derive a contradiction. Conclude that $c=0$. $\endgroup$ – Mastrem Jun 10 at 19:18

Your Answer

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

Browse other questions tagged or ask your own question.