# Suppose x_n in R, x_1=1, 2x_{n+1}=x_n+3/x_n. Then show that limxn exists and find its value.

Is this analysis problem or discrete math problem?
Suppose ${x}_{n}\in \mathbb{R},{x}_{1}=1,2{x}_{n+1}={x}_{n}+3/{x}_{n}$. Then show that limxn exists and find its value.
So is this problem (real) analysis problem or a discrete math one?
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Step 1
First, we will show that $\left\{{x}_{n}\right\}$ is strictly decreasing and bounded (not counting the first term). To do this, we must show that ${x}_{n+1}=\frac{{x}_{n}}{2}+\frac{3}{2{x}_{n}}<{x}_{n},$, which is equivalent to ${x}_{n}>\frac{3}{{x}_{n}},$, i.e. ${x}_{n}>\sqrt{3}.$ This last statement we can prove easily (for $n\ge 2$). By AM-GM, ${x}_{n+1}=\frac{{x}_{n}}{2}+\frac{3}{2{x}_{n}}\ge \sqrt{3},$, with equality iff ${x}_{n}=\sqrt{3}.$. Since ${x}_{n}\ne \sqrt{3}$ for $n=2,$, equality never occurs, i.e. ${x}_{n}>\sqrt{3}$ for all $n\ge 2,$, from which we have $\sqrt{3}<{x}_{n+1}<{x}_{n},$, as noted above.
Step 2
By the monotone convergence theorem, $\left\{{x}_{n}\right\}$ has a limit, say x. Then, by the given, $2x=x+\frac{3}{x},$, from which $x=\sqrt{3}.$.