A recursive sequence (2)

Re-index the sequence as $\begin{cases} a_{\color{#D61F06}0} = 2017 \\ a_{\color{#D61F06}1} = 2117 \\ a_{n+2} = \frac 25 a_{n+1} + \frac 35 a_n, \quad n \ge {\color{#D61F06}0} \end{cases}$

The characteristic of the linear recurrence relation $a_{n+2} = \frac 25 a_{n+1} + \frac 35 a_n$ is as follows:

$\begin{aligned} r^2 - \frac 25 r - \frac 35 & = 0 \\ 5r^2 - 2 r - 3 & = 0 \\ (5r+3)(r-1) & = 0 \\ \implies r & = - \frac 35, \ 1 \end{aligned}$

$\begin{aligned} \implies a_n & = c_1 + c_2 \left(-\frac 35\right)^n \\ a_0 & = c_1 + c_2 = 2017 \\ a_1 & = c_1 - \frac 35 c_2 = 2117 \\ \implies c_1 & = 2079.5 \\ c_2 & = -62.5 \\ \implies a_n & = 2079.5 + (-1)^{n+1} 62.5 \times 0.6^n \end{aligned}$

Therefore, $\displaystyle \lim_{n \to \infty} a_n = \lim_{n \to \infty} \left(2079.5 + (-1)^{n+1} 62.5 \times 0.6^n\right) = \boxed{2079.5}$

From the condition, $a_{k+2}-a_{k+1}=\frac{-3}{5}\left(a_{k+1}-a_k\right)$ . Hence $\{a_{k+2}-a_{k+1}\}$ is a geometry progression. Now $a_{k+2}-a_{k+1}=\left(\frac{-3}{5}\right)^k\left(a_{2}-a_1\right) =100\left(\frac{-3}{5}\right)^k$ . Take the summation both sides from $k=1$ to $k=n$ , we obtain $a_{n+2}-a_2 = 100\left( \frac{-3}{5} + \left(\frac{-3}{5}\right)^2+\left(\frac{-3}{5}\right)^3 +\ldots+ \left(\frac{-3}{5}\right)^n \right)$ .

Now $\displaystyle \lim_{n \to \infty } a_n = \lim_{n \to \infty } a_{n+2}=a_2 + 100\left(\frac{\frac{-3}{5} }{1-\left(\frac{-3}{5}\right)}\right) =\boxed{2079.5}$ .