# How to solve a recursive equation with fraction structure

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 $$a(K) + b(K) = K$$,

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

• 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 $c=0$. – Mastrem Jun 10 at 19:18