If $u_{n+1} = u_n^2 - u_n + 1$ and $u_0=a>1$, show that $\sum\frac{1}{u_n} \rightarrow \frac{1}{a-1}$ If $(u_n)\in\mathbb{R}^\mathbb{N},u_{n+1} = u_n^2 - u_n + 1$ and $u_0=a>1$, show that $\sum\dfrac{1}{u_n} \rightarrow \dfrac{1}{a-1}$.
I don't even know how to prove that the series converges
 A: The key observation, which kills theproblem quite instantly, is the following: for every $n$ it holds the relation $$\frac{1}{u_n} = \frac{1}{u_n - 1} - \frac{1}{u_{n + 1} - 1}$$ Indeed we have $$\frac{1}{u_n - 1} - \frac{1}{u_{n + 1} - 1} = \frac{1}{u_n - 1} - \frac{1}{u_n(u_n - 1)} = \frac{u_n - 1}{u_n(u_n - 1)} = \frac{1}{u_n}$$
Hence we have a telescopic sum: $$\sum_{n = 0}^{N} \frac{1}{u_n} = \sum_{n = 0}^N \left(\frac{1}{u_n - 1} - \frac{1}{u_{n + 1} - 1}\right) = \\ = \left(\frac{1}{u_0 - 1} - \frac{1}{u_1 - 1}\right) + \left(\frac{1}{u_1 - 1} - \frac{1}{u_2 - 1}\right) + \cdots + \left(\frac{1}{u_N - 1} - \frac{1}{u_{N + 1} - 1}\right) = \\ = \frac{1}{a - 1} + \frac{1}{u_{N + 1} - 1}$$ Now it is pretty easy to prove that $\displaystyle \lim_{n \rightarrow \infty} u_n = \infty$ and thus $$\lim_{N \rightarrow \infty} \sum_{n = 0}^N \frac{1}{u_i} = \lim_{N \rightarrow \infty} \left(\frac{1}{a - 1} + \frac{1}{u_{N + 1} - 1}\right) = \frac{1}{a - 1}$$
So the series converges to $\dfrac{1}{a - 1}$.
A: Going the other way,
if
$\dfrac1{u_n}
=\dfrac1{u_n-a}-\dfrac1{u_{n+1}-a}
$,
so that
$\sum\dfrac1{u_n}$
converges via telescoping,
then,
simplifying,
$u_n^2-au_n+a^2
=au_{n+1}
$.
This is the case
$a=1$.
Of course the seemingly
more general case
reduces to the case
$a=1$
by replacing $u_n$
by
$\dfrac{u_n}{a}
$.
