Supremum of a sequence: $x_n = c_1 + c_2 \sum_{i=1}^{n-1} x_i $

Question

I am dealing with a sequence

$$ x_n = c_1 + c_2 \sum_{i=1}^{n-1} x_i $$

where $x_1 > 0$ and $c_1, c_2 > 0$ are constants. I am wondering whether one can find another series, call it $y_n$, that can be explicitly expressed in terms of $n$ and satisfies $x_n \leq y_n$ for each $n$.

I tried to rewrite the above recursive formula as $$ (1 + c_2) x_n = c_1 + c_2 S_n $$ where $S_n = \sum_{i=1}^n x_i$, and obtained $$ (c_2 n - 1) S_n = c_2 \sum_{i=1}^n i x_i - c_1 n $$ I am struggling to find a way forward. Do you have any suggestions?

Answer 1

You may observe that $$ x_n = c_1 + c_2 \sum_{i=1}^{n-1} x_i \tag1 $$ gives $$ x_{n+1} = c_1 + c_2 \sum_{i=1}^{n} x_i. \tag2 $$ By making $(2)-(1)$, you just get $$ x_{n+1}-x_n=c_2x_n \tag3 $$ or

$$ x_{n+1}=(1+c_2)\:x_n \tag4 $$

which is easier to deal with.