If $a_{n+1}=1+\frac{1}{a_{1}+a_{2}+\cdots+a_{n}-1}$ then $0define the sequence $\{a_{n}\}$ and such
$$a_{1}=a,a_{n+1}=1+\dfrac{1}{a_{1}+a_{2}+\cdots+a_{n}-1},n\ge 1$$
Find the all real number $a>0$,such 
$$0<a_{n}<1,n\ge 2$$
My try: since $$a_{2}=1+\dfrac{1}{a_{1}-1}=1+\dfrac{1}{a-1}$$
then
$$0<1+\dfrac{1}{a-1}<1$$
so
$$0<a<1$$
and $$a_{3}=1+\dfrac{1}{a_{1}+a_{2}-1}=1+\dfrac{1}{a+\dfrac{a}{a-1}-1}=\dfrac{a^2}{a^2-a+1}$$
if $$a_{3}\in(0,1)\Longrightarrow 0<a<1$$
so maybe I guess if $0<a<1$,then $a_{n}\in(0,1),n\ge 2$
 A: We will prove that the necessary and sufficient condition on $a$ to have $0<a_n<1$ for all $n\geq2$ is ''($a<0$)''.
Indeed the inequality $a_2\in(0,1)$ is equivalent to $0<\frac{a}{a-1}<$ ind this is equivalent
to $(a<0)$. So, this is a necessary condition.
Now, suppose that $a<0$. Let $s_n=-1+\sum_{k=1}^na_k$, the sequence $\{s_n\}$ is defined inductively by
$$ s_1=a-1<-1,\qquad, s_{n+1}=f(s_n),\quad\hbox{with $f(x)=1+x+\frac{1}{x}$}$$
$f$ is strictly increasing on $(-\infty,-1)$, and $f((-\infty,-1))=(-\infty,-1)$, so if
$s_n\in(-\infty,-1)$ then $s_{n+1}\in (-\infty,-1)$ also, This shows by induction that,
$s_n<-1$ for every $n\geq 1$ since $s_1=a-1<-1$.
It follows that $a_{n+1}=f(s_n)-s_n=\frac{1+s_n}{s_n}\in(0,1)$ for every $n\geq1$ which is the desired conclusion.
A: First, we try to simplify the given sequence:
$$a_{n+1} - 1 = \frac{1}{a_1 + \dots + a_n - 1}$$
$$a_1 + \dots + a_n - 1 = \frac{1}{a_{n+1} - 1}$$
$$a_1 + \dots + a_n = 1 + \frac{1}{a_{n+1} - 1}$$
Now, we do some subtraction for $n\ge2$:
$$(a_1 + \dots + a_{n-1} + a_n) - (a_1 + \dots + a_{n-1}) = \left(1 + \frac{1}{a_{n+1} - 1}\right) - \left(1 + \frac{1}{a_{n} - 1}\right)$$
$$a_n = \frac{1}{a_{n+1} - 1} - \frac{1}{a_n - 1}$$
$$\frac{1}{a_{n+1} - 1} = \frac{a_n^2 - a_n + 1}{a_n - 1}$$
$$a_{n+1} - 1 = \frac{a_n - 1}{a_n^2 - a_n + 1}$$
$$a_{n+1} = \frac{a_n^2}{a_n^2 - a_n + 1}$$
Now, 
$$0 < a_{n+1} < 1$$
$$\iff 0 <  \frac{a_n^2}{a_n^2 - a_n + 1} < 1$$
But $a_n^2 - a_n + 1$ is always positive for all real $a_n$ (consider its discriminant). Hence, we can multiply throughout to get:
$$0 < a_n^2 < a_n^2 - a_n + 1$$
From here, we can deduce that $a_n \not = 0, a_n < 1 \implies 0 < a_{n+1} < 1 \implies 0 < a_{n+2} < 1 \implies \dots$. As such, $a_2 \not = 0, a_2 < 1 \implies 0 < a_n < 1$ for all positive integers $n \ge 2$. 
But $0<a_2<1 \implies a_2 \not = 0, a_2 < 1$. So it suffices to find the values of $a_2$ for which $0< a_2 < 1$. But $ a_2 = 1 + \frac{1}{a - 1}$. Hence,
$$0< 1 + \frac{1}{a - 1} < 1$$
$$-1 < \frac{1}{a-1} < 0$$
From $\frac{1}{a-1} < 0$, we have $$a - 1 < 0 \implies a < 1$$ 
We also have (from $-1 < \frac{1}{a-1}$):
$$-a + 1 > 1 \implies a < 0$$
To conclude, the valid range of $a$ is given by $a < 0$.
