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### Define $$x_1 := 2$$ and $$x_n := \tfrac{7}{9}\left(x_{n-1}+5\right)$$. Prove that $$(x_n)$$ converges and find the limit.

Where do you start? We compute the first few terms in the sequence to find our whether the sequence is increasing or decreasing, then we assume the sequence converges to find out what the limit is before we try to prove it.

We compute that $$x_2\approx 5.4,\ x_3 \approx 8.1, x_4 \approx 10.2, \dots ,$$ so the sequence seems to be increasing. Passing to the limit we learn that $$x=\tfrac{7}{9}(x+5)\Rightarrow x=35/2.$$ Now that we're reasonably sure that the sequence is increasing and converging to 35/2, we are ready to prove it!

Try writing the proof yourself, then click on the box below to see a finished proof.

Show the Proof

Proof: We will show that $$(x_n)$$ is bounded above by 18 and increasing. First, $$x_1 = 2 \le 18$$, so assume that $$x_k \le 18$$ for some $$k\in\mathbb N$$. Then $$x_k + 5\le 23$$ $$\Rightarrow$$ $$\tfrac{7}{9}\left(x_{k}+5\right) \le 161/9$$ $$\Rightarrow$$ $$x_{k+1} \le 162/9=18$$. Thus $$x_n \le 18$$ for all $$n\in \mathbb N$$.

Because $$x_1 = 2$$ and $$x_2 =49/9 > 2$$ we assume that $$x_{k-1} \le x_k$$ for some $$k\in\mathbb N$$. Then $$x_{k-1} + 5\le x_k + 5$$ $$\Rightarrow$$ $$\tfrac{7}{9}\left(x_{k-1}+5\right) \le \tfrac{7}{9}\left(x_{k}+5\right)$$ $$\Rightarrow$$ $$x_{k} \le x_{k+1}$$. Thus $$x_n \le x_{n+1}$$ for all $$n\in \mathbb N$$, so $$(x_n)$$ is increasing.

We have shown that $$(x_n)$$ is increasing and bounded above, so by the Monotone Convergence Theorem, $$(x_n)$$ converges to a real number $$x$$. Passing to the limit, we learn that $$x=\tfrac{7}{9}(x+5)\Rightarrow x=35/9.$$