Given the sequence defined by the recursion $x_{n+1}=x_{n}^{2}-x_{n}+1$, construct the infinite series $\sum_{i=1}^{\infty} \frac{1}{x_{i}}$. What is the sum of this series if $a$ ) $x_{1}=1 / 2 ; b$ ) $x_{1}=2$ ?
Solution. The sequence is given by the recursion $x_{n+1}=f\left(x_{n}\right)$, where $f(x)=x^{2}-x+1$. The graph of the function $f$ is shown in the figure, and we have also drawn the line $y=x$, which is tangent to the curve at the point $(1 ; 1)$. From the graph, it is clear that $f(x) \geq \frac{3}{4}$, so if $x_{1} \neq 0$, then every term of the $\frac{1}{x_{i}}$ sequence is well-defined.

![](https://cdn.mathpix.com/cropped/2024_05_02_d25bf9262e50aebcb4e7g-1.jpg?height=713&width=569&top_left_y=337&top_left_x=775)

If $0 < x_{1} < 1$, then $x_{2} > 1$, and if $x_{1} > 1$, then $x_{n+1} > x_{n}$. In the latter case, we have:

\[
\begin{aligned}
x_{n+1}-1 & =x_{n}^{2}-x_{n} = x_{n}(x_{n}-1) \\
& = x_{n} x_{n-1}(x_{n-1}-1) = \cdots = x_{n} x_{n-1} \cdots x_{1}(x_{1}-1) \\
& > n(x_{1}-1)
\end{aligned}
\]

thus, the sequence $x_{n}$ tends to infinity, and therefore the limit of the sequence $\frac{1}{x_{n+1}-1}$ is zero.

The limit of the sequence $S_{n}$ is therefore $\frac{1}{x_{1}-1}$ when $x_{1}>1$, and if $x_{1}=2$, the sum of the infinite series $\sum_{i=1}^{\infty} \frac{1}{x_{i}}$ is 1.

Remarks. 1. If $x_{1}<0$, then the second term of the sequence is greater than 1, and thus, according to the above, the sum of the infinite series $\sum_{i=1}^{\infty} \frac{1}{x_{i}}$ is also $\frac{1}{x_{1}-1}$ in this case.

2. It can be easily shown that if $x_{1}$ is not 0 or 1, then $\left(x_{1}-1\right) \sum_{i=1}^{n} \frac{1}{x_{i}}+\prod_{i=1}^{n} \frac{1}{x_{i}}=1$. The questions of the problem can also be answered quickly using this identity.
3. In the case where $x_{1}=2$, the sequence $x_{n}$ has interesting properties. First, it can be easily shown that in this case, the sequence elements are calculated by the recursion $x_{n+1}=1+x_{1} x_{2} \cdots x_{n}$: $x_{1}=2, x_{2}=3, x_{3}=7$, $x_{4}=43, \ldots$, and secondly, the following statement holds for this sequence: if the sum of the reciprocals of $n$ positive integers is less than 1, then this sum is at most $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots+\frac{1}{x_{n}}$. If one thinks about it, it is not even obvious that among the sums of reciprocals of $n$ terms that are less than 1, there is a largest one at all...