Question

Let $a_0 = 0, a_1 = 2$ and the sequence defined recursively by $a_{n+1} = \sqrt{2-\frac{a_{n-1}}{a_n}}$ for $n \ge 1$. Find

$\lim_{n \to \infty} 2^n \cdot a_n$

• A. 4π
• B. 3π
• C. 2π
• D. π

Answer 1

Answer: (C)

2π

Answer 2

The given sequence is defined by $a_0 = 0$, $a_1 = 2$, and $a_{n+1} = \sqrt{2-\frac{a_{n-1}}{a_n}}$ for $n \ge 1$.

First, let's compute the first few terms of the sequence: $a_0 = 0$ $a_1 = 2$ For $n=1$: $a_2 = \sqrt{2 - \frac{a_0}{a_1}} = \sqrt{2 - \frac{0}{2}} = \sqrt{2}$. For $n=2$: $a_3 = \sqrt{2 - \frac{a_1}{a_2}} = \sqrt{2 - \frac{2}{\sqrt{2}}} = \sqrt{2 - \sqrt{2}}$.

We observe a pattern related to trigonometric functions. Let's try to express $a_n$ in the form $2\sin(\theta_n)$: $a_1 = 2 = 2\sin(\pi/2)$, so $\theta_1 = \pi/2$. $a_2 = \sqrt{2} = 2(\sqrt{2}/2) = 2\sin(\pi/4)$, so $\theta_2 = \pi/4$. $a_3 = \sqrt{2-\sqrt{2}} = 2\sqrt{\frac{2-\sqrt{2}}{4}} = 2\sqrt{\frac{1-\sqrt{2}/2}{2}} = 2\sqrt{\frac{1-\cos(\pi/4)}{2}}$. Using the half-angle identity $\sin(x/2) = \sqrt{(1-\cos x)/2}$, we get $a_3 = 2\sin(\pi/8)$, so $\theta_3 = \pi/8$.

The pattern suggests that $a_n = 2\sin\left(\frac{\pi}{2^n}\right)$. Let's verify this formula for $a_0$: $a_0 = 2\sin\left(\frac{\pi}{2^0}\right) = 2\sin(\pi) = 0$, which matches the given value.

Now, we verify if this formula satisfies the recurrence relation $a_{n+1} = \sqrt{2-\frac{a_{n-1}}{a_n}}$. Substitute $a_k = 2\sin\left(\frac{\pi}{2^k}\right)$ into the right-hand side (RHS) of the recurrence: RHS $= \sqrt{2 - \frac{2\sin(\frac{\pi}{2^{n-1}})}{2\sin(\frac{\pi}{2^n})}} = \sqrt{2 - \frac{\sin(\frac{\pi}{2^{n-1}})}{\sin(\frac{\pi}{2^n})}}$. Let $x = \frac{\pi}{2^n}$. Then $\frac{\pi}{2^{n-1}} = 2x$ and $\frac{\pi}{2^{n+1}} = x/2$. RHS $= \sqrt{2 - \frac{\sin(2x)}{\sin x}}$. Using the double-angle identity $\sin(2x) = 2\sin x \cos x$: RHS $= \sqrt{2 - \frac{2\sin x \cos x}{\sin x}}$. Since $n \ge 1$, $x = \frac{\pi}{2^n}$ is in $(0, \pi/2]$, so $\sin x \ne 0$. Thus, we can simplify: RHS $= \sqrt{2 - 2\cos x} = \sqrt{2(1-\cos x)}$. Using the half-angle identity $1-\cos x = 2\sin^2(x/2)$: RHS $= \sqrt{2(2\sin^2(x/2))} = \sqrt{4\sin^2(x/2)} = 2|\sin(x/2)|$. Since $x = \frac{\pi}{2^n}$ and $n \ge 1$, $x/2 = \frac{\pi}{2^{n+1}}$ is in $(0, \pi/4]$. In this range, $\sin(x/2)$ is positive. So, RHS $= 2\sin(x/2) = 2\sin\left(\frac{\pi}{2^{n+1}}\right)$. This is exactly $a_{n+1}$ according to our formula. Therefore, $a_n = 2\sin\left(\frac{\pi}{2^n}\right)$ is the correct general term for the sequence.

Finally, we need to find the limit: $\lim_{n \to \infty} 2^n \cdot a_n$. Substitute $a_n = 2\sin\left(\frac{\pi}{2^n}\right)$: $\lim_{n \to \infty} 2^n \cdot 2\sin\left(\frac{\pi}{2^n}\right)$. Let $y = \frac{\pi}{2^n}$. As $n \to \infty$, $2^n \to \infty$, so $y \to 0$. The expression becomes: $\lim_{y \to 0} \frac{\pi}{y} \cdot 2\sin(y) = 2\pi \lim_{y \to 0} \frac{\sin y}{y}$. Using the standard limit $\lim_{y \to 0} \frac{\sin y}{y} = 1$: The limit is $2\pi \cdot 1 = 2\pi$.

Explanation of the solution:

1. Calculate Initial Terms: Compute $a_2 = \sqrt{2}$ and $a_3 = \sqrt{2-\sqrt{2}}$ using the given recurrence and initial values.
2. Identify Pattern: Observe that $a_1=2\sin(\pi/2)$, $a_2=2\sin(\pi/4)$, $a_3=2\sin(\pi/8)$. This suggests the general term $a_n = 2\sin(\pi/2^n)$. Verify this for $a_0$.
3. Verify Recurrence: Substitute $a_k = 2\sin(\pi/2^k)$ into the given recurrence relation $a_{n+1} = \sqrt{2-\frac{a_{n-1}}{a_n}}$. Use trigonometric identities $\sin(2x) = 2\sin x \cos x$ and $1-\cos x = 2\sin^2(x/2)$ to show that the RHS simplifies to $2\sin(\pi/2^{n+1})$, which is $a_{n+1}$.
4. Evaluate Limit: Substitute $a_n = 2\sin(\pi/2^n)$ into the expression $\lim_{n \to \infty} 2^n \cdot a_n$. Let $y = \pi/2^n$. As $n \to \infty$, $y \to 0$. The limit becomes $\lim_{y \to 0} \frac{\pi}{y} \cdot 2\sin y = 2\pi \lim_{y \to 0} \frac{\sin y}{y}$.
5. Apply Standard Limit: Use $\lim_{y \to 0} \frac{\sin y}{y} = 1$ to find the final limit as $2\pi$.