Let $\alpha, \beta$ be real numbers. Find all convergent sequences $(a_n)_{n \geq 1}$ satisfying$$\alpha(a_1 + \cdots + a_n) + \beta(a_1 \cdot \ldots \cdot a_n) = 1$$for all $n \geq 1$.
AttemptNote that $(\alpha+\beta)a_1=1$ hence we must have $\alpha+\beta\not=0$.If $\alpha=0$ then $\beta p_n=1$ for all $n$ hence $a_2=a_3=\cdots=1$ Also $a_1=1/\beta$. From now on $\alpha\not=0$.If $\beta=0$ then $\alpha s_n=1$ for all $n$ hence $a_2=a_3=\cdots=0$ and $a_1=1/\alpha$. From now on $\alpha\beta\not=0$.
We have $\alpha s_n+\beta p_n=\alpha s_{n+1}+\beta p_{n+1}\iff \beta p_n(1-a_{n+1})=\alpha a_{n+1}\iff \beta p_n=(\alpha+\beta p_n)a_{n+1}$.Note that $\alpha+\beta p_n\not=0$ for all $n$ hence we may divide and find $a_{n+1}={{\beta p_n}\over {\beta p_n+\alpha}}$. This shows that the sequence is unique, if it exists. Also note with induction that $a_n\not=0$ for all $n$.Multiplying by $p_n$ we find $p_{n+1}={{\beta p_n^2}\over {\beta p_n+\alpha}}$ and with $q_n:=\beta p_n$ we find $q_{n+1}={{q_n^2}\over {q_n+\alpha}}$. This is a recursion where $q_1={\beta\over {\alpha+\beta}}$ depends on $\alpha,\beta$ and where the recursion function $f(x)={{x^2}\over {x+\alpha}}$ only depends on $\alpha$. Then $a_n$ converges iff $q_n$ converges or if $|q_n|\rightarrow \infty$. If $q_n$ converges then $p_n$ also and we must have $p_n\rightarrow 0$ since otherwise $a_n\rightarrow 1 \implies |q_n|\rightarrow \infty$.Hence we must have $q_n\rightarrow 0$ or $|q_n|\rightarrow \infty$.
Hope this attempt helps you