Converging series

For any $x_0$ the series converges to be consistent with $x_\infty = \frac{x_\infty}{8} + 3$

$\implies$ $x_\infty = \frac{24}{7}$

$24+7 = \boxed{31}$

Relevant wiki: Linear Recurrence Relations

Assuming $\displaystyle \lim_{n \to \infty} x_n$ exists and let it be $L$ . Then we have:

$\begin{aligned} x_n & =\frac {x_{n-1}}8+3 \\ \lim_{n \to \infty} x_n & = \lim_{n \to \infty} \frac {x_{n-1}}8+3 \\ L & = \frac L8+3 \\ \frac 78 L & = 3 \\ \implies L & = \frac {24}7 \end{aligned}$

$\implies a+b=24+7=\boxed {31}$

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Solution without assuming the limit exists.

$\begin{aligned} x_n & = \frac {x_{n-1}}8 + 3 & \small \color{#3D99F6}{\text{To make it a linear recurrence relation, let }y_n = x_n - u} \\ x_n - u & = \frac {x_{n-1}-u}8 \\ x_n & = \frac {x_{n-1}}8 + u - \frac u8 \\ \implies \frac 78u & = 3 \\ u & = \frac {24}7 \end{aligned}$

Now, we have:

$\begin{aligned} y_n & = \frac {y_{n-1}}8 & \small \color{#3D99F6}{\text{The characteristic equation is }r=\frac 18} \\ \implies y_n & = \frac a{8^n} \\ x_n & = \frac a{8^n} + \frac {24}7 \\ \implies x_0 & = a + \frac {24}7 \\ a & = x_0 - \frac {24}7 \\ \implies x_n & = \frac {x_0 - \frac {24}7}{8^n} + \frac {24}7 \\ \implies x_\infty & = \frac {24}7 \end{aligned}$

$\implies a+b = 24+7 = \boxed{31}$