Task B-4.3. Let $f_{n}(x)=x^{n+1}+x^{n}$, for $n \in \mathbb{N}, x \in \mathbb{R}$. For which real numbers $x$, will the infinite sum $f_{1}(x)+f_{2}(x)+f_{3}(x)+\ldots$ have a value in the interval $\left\langle 0, \frac{2}{3}\right]$?
## Solution.

Dani is a sequence of functions forming a geometric sequence, where the first term is $a_{1}=f_{1}(x)=x^{2}+x$, and the common ratio is $q=x$. Then the infinite sum $f_{1}(x)+f_{2}(x)+f_{3}(x)+f_{4}(x)+\ldots$ is a geometric series. This geometric series will converge if $|q|<1$, that is, if $|x|<1, x \neq 0$. In this case, the sum is

$$
f_{1}(x)+f_{2}(x)+f_{3}(x)+f_{4}(x)+\ldots=\frac{a_{1}}{1-q}=\frac{x^{2}+x}{1-x}
$$

If $x=0$, the desired sum is 0, which does not fall within the given interval.

We need to determine all $x \in \mathbb{R}, x \neq 0$ for which the system of inequalities holds

$$
0<\frac{x^{2}+x}{1-x} \leq \frac{2}{3}
$$

The solutions to the inequalities are

$$
\begin{aligned}
& \frac{x^{2}+x}{1-x} \leq \frac{2}{3}, \text { for } x \in\left[-2, \frac{1}{3}\right] \cup\langle 1, \infty\rangle \\
& 0<\frac{x^{2}+x}{1-x}, \text { for } x \in\langle-\infty,-1\rangle \cup\langle 0,1\rangle
\end{aligned}
$$

The intersection of these solutions, considering the convergence condition that $|x|<1$, gives the final solution

$$
x \in\left\langle 0, \frac{1}{3}\right]
$$