Let $x_1 = 2$ and $x_{n+1} = 1 + sqrt{x_n - 1}$ for $n in mathbb{N}$. Show that ${x_n}$ is decreasing and is bounded below by 2. Find its limit.
Added by Kevin M.
Step 1
To do this, we need to show that x_n > x_{n+1} for all n in N. We have x_{n+1} = 1 + (x_n)^{1/2}. Since x_1 = 2, we know that x_n ≥ 2 for all n in N. Therefore, (x_n)^{1/2} ≥ (2)^{1/2} = √2. Now, let's consider the difference between x_n and x_{n+1}: x_n - Show more…
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