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Does the following:

$$ \begin{align} x_0 & = a \\ x_1 & = x_0 + \frac{1}{1}(x_0 + x_0(c - 1)) \\ x_2 & = x_1 + \frac{1}{2}(b + x_1(c - 1)) \\ x_3 & = x_2 + \frac{1}{3}(b + x_2(c - 1)) \\ & {}\,\vdots \\ x_n & = x_{n-1} + \frac{1}{n}(b + x_{n-1}(c - 1)) \end{align} $$

where $x_{0}$ is switched to b in the $x_{2}$ term onwards, converge to:

$$\frac{b}{1 - c}\quad\text{?}$$

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up vote 5 down vote accepted

First, the constant $c$ is slightly misleading. Why not take $d := 1-c$ so that: $$x_{n+1} = \left(1-\frac{d}{n}\right) x_{n} + \frac{b}{n}$$ Let $y_n := x_n - t$ for some $t$ that we'll establish in a second. Then: $$ y_{n+1} + t = \left(1-\frac{d}{n}\right) y_{n} + \left(1-\frac{d}{n}\right)t + \frac{b}{n}$$ which is more naturally written as: $$ y_{n+1} = \left(1-\frac{d}{n}\right) y_{n} + \frac{b-td}{n}$$ Thus, let $t := \frac{b}{d}$, so that we have: $$ y_{n+1} = \left(1-\frac{d}{n}\right) y_{n} $$ This allows you to compute $y_n$ almost explicitly: $$ y_{n+1} = y_k\prod_{m=k}^n \left(1-\frac{d}{m}\right) $$ where $k$ is reasonably small (say, $k=3$). It is a classical theorem in analysis that because $\sum_m \frac{d}{m}$ diverges to $+\infty$, the product $\prod_{m=k}^n \left(1-\frac{d}{m}\right) $ converges to $0$ as $n \to \infty$. Thus, we have $y_n \to 0$ as $n \to \infty$. We may now recover $x_n$ and find its limit: $x_n = y_n + t \to t = \frac{b}{1-c}$. Hence, the conjectured convergence is true. And it doesn't really matter is anything changes at the initial steps in the induction.

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ah ok yes. This was the best proof i saw in days , great job. – sigmatau May 17 '13 at 21:22
@AmireBendjeddou: Thank you, it is a very kind thing of you to write. I am glad I could be helpful. – Jakub Konieczny May 17 '13 at 21:36

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