Question

A sequence of rational numbers is described as follows:
1/1,3/2,7/5,17/12,…,a/b,(a + 2b)/(a + b),…
Here the numerators form one sequence, the denominators form a second sequence, and their ratios form a third sequence. Let xₙ and yₙ be, respectively, the numerator and the denominator of the nᵗʰ fraction rₙ = xₙ / yₙ.
b. The fractions rₙ = xₙ / yₙ approach a limit as n increases. What is that limit? (Hint: Use part (a) to show that rₙ² − 2 = ±(1 / yₙ)² and that yₙ is not less than n.)

Accepted Answer

First, recognize that the sequence of fractions is defined recursively by the relation: $$r_{n+1} = \frac{a + 2b}{a + b}$$, where $$r_n = \frac{x_n}{y_n} = \frac{a}{b}. $$This means $$r_{n+1} = \frac{x_n + 2 y_n}{x_n + y_n}. $$Express $$r_{n+1} in $$terms of $$r_n by $$dividing numerator and denominator by $$y_n$$: $$r_{n+1} = \frac{\frac{x_n}{y_n} + 2}{\frac{x_n}{y_n} + 1} = \frac{r_n + 2}{r_n + 1}. $$This gives a recursive formula for $$r_n. To $$find the limit $$L of$$ $$r_n as$$ $$n$$ approaches infinity, assume the limit exists and satisfies $$L = \frac{L + 2}{L + 1}. $$Solve this equation for $$L by $$multiplying both sides by $$(L + 1) to $$get $$L(L + 1) = L + 2. $$Simplify the equation: $$L^2 + L = L + 2$, which reduces to L^2$ = 2. $$Therefore, the possible limits are $$L = \sqrt{2} or$$ $$L = -\sqrt{2}. $$Since the sequence terms are positive, the limit is $$L = \sqrt{2}. $$Use the hint from part (a) that $$r_n^2$ - 2 = \pm \frac{1}{y_n^2$}$$ and that $$y_n$$ grows at least as fast as $$n$$, so $$\frac{1}{y_n^2$} 	o 0 as$$ $$n 	o \infty. $$This confirms that $$r_n^2$ approaches 2, reinforcing that r_n$ approaches $$\sqrt{2}$$.

비슷한 문제에 대한 검증된 영상 답변:

주요 개념

Recursive Sequences and Their Limits

Limits of Sequences and Convergence

Using Algebraic Identities to Find Limits