Convergence of sequence 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:
$$\frac{b}{1 - c}\quad\text{?}$$
 A: 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. 
