[College Calc 01-02. Limits of Sequences] #06

Let us suppose that $a_n$ converges to $\alpha.$ Then, we easily see that

$\alpha = \frac{2}{3}\alpha+2\implies \alpha = 6.$

Then, let $a_n = b_n + 6$ so that

$b_{n+1} + 6 = \frac{2}{3}(b_n+6) + 2\implies b_{n+1} = \frac{2}{3}b_n\implies b_n = k\left(\frac{2}{3}\right)^n.$

Substituting back, we have

$a_n = k\left(\frac{2}{3}\right)^n + 6\implies a_0 = k+6 = 1\implies k=-5.$

Using the sum formula for geometric sequences ,

$\sum_{k=0}^{n} a_k = \sum_{k=0}^{n} \left[-5\left(\frac{2}{3}\right)^k + 6\right] = -5\cdot \frac{1-\left(\frac{2}{3}\right)^{n+1}}{1-\frac{2}{3}} + 6(n+1),$

and thus

$\lim_{n\to\infty}\left(pn - \sum_{k=0}^{n}a_k\right) = \lim_{n\to\infty} (p-6)n + 9 = q.$

Clearly if the limit is to converge, $p=6$ such that $q=9.$

$\therefore~p+q=6+9 = \boxed{15}.$

Given that

$\begin{aligned} a_{n+1} & = \frac 23 a_n + 2 \\ a_{n+1} - 6 & = \frac 23 \left(a_n - 6 \right) \end{aligned}$

Let $b_n = a_n - 6$ . Then $b_{n+1} = \dfrac 23 b_n$ and the characteristic equation of the linear recurrent relations is $r = \dfrac 23$ and $b_n = c \left(\dfrac 23\right)^n$ , where $c$ is a constant. Since $b_0 = 1 - 6 = - 5 = c$ , $b_n = a_n - 6 = -5 \left(\dfrac 23\right)^n$ , and

$\begin{aligned} a_k & = 6 - 5\left(\dfrac 23\right)^k \\ \sum_{k=0}^n a_k & = \sum_{k=0}^n \left(6-5\left(\dfrac 23\right)^k\right) = 6(n+1) - 5 \sum_{k=0}^n \left(\frac 23\right)^k \\ 6n - \sum_{k=0}^n a_k & = 5 \sum_{k=0}^n \left(\frac 23\right)^k - 6 \\ \implies \lim_{n \to \infty}\left(6n - \sum_{k=0}^n a_k \right) & = 5 \cdot \frac 1{1-\frac 23} - 6 = 9 \end{aligned}$

Therefore $p+q = 6 + 9 = \boxed{15}$ .

$a_{n+1} = \frac 23 a_n + 2 \\ a_{n+1} - 6 = \frac 23 a_n + 2 - 6 = \frac 23 (a_n - 6)$

let $X_n = a_n - 6$ , then $X_0 = a_0 - 6 = -5$ , $X_{n+1} = \frac 23 X_n$

$\sum\limits_{i=0}^n X_i = \sum\limits_{i=0}^n (a_i - 6) \\ \frac {X_0 \cdot (1-(\frac 23)^n)}{1 - \frac 23} = \sum\limits_{i=0}^n a_i - 6(n+1) \\ -15(1 - (\frac 23)^n) = \sum\limits_{i=0}^n a_i - 6(n+1)$

$\lim\limits_{n \to \infty} \sum\limits_{i=0}^n X_i = \lim\limits_{n \to \infty}(\sum\limits_{i=0}^n a_i - 6(n+1)) \\ -15 = \lim\limits_{n \to \infty}(\sum\limits_{i=0}^n a_i - 6(n+1)) \\ \lim\limits_{n \to \infty}(6n -\sum\limits_{i=0}^n a_i) = 9 \\ \therefore p + q = 6 + 9 = 15$

Let's start from the relation

$a_{n+1}=2/3a_n+2 \implies a_{n+1}-a_n=2-a_n/3$

Using this, we have

$\sum_{k=0}^n (2-a_k/3) = \sum_{k=0}^n (a_{k+1}-a_k) \implies 2(n+1) - \frac{1}{3}\sum_{k=0}^n a_k = a_{n+1}-a_0 \implies \sum_{k=0}^n a_k = 6n-2a_n+3$

From this, since we want the limit to be finite, we get p=6; let's calculate now the limit of $a_n$ : since $f(n)=2/3n +2$ is a contraction, we have that $a_n$ converges to the fixed point of $f(n)$ , which is 6. Now, we have

$q= \lim_{n} (pn-\sum_{k=0}^n a_k) = \lim_n (2a_n-3) = 9$

Therefore, $p+q=15$