Problem
The sequence (an)∞n=1 is given by recurrence relation:
- a1=√2,
- an+1=√2+an.
Evaluate the limit lim.
Solution
- Show that the sequence (a_n)_{n=1}^\infty is monotonic. The statement V(n): a_n < a_{n+1} holds for n = 1, that is \sqrt2 < \sqrt{2+\sqrt2}. Let us assume the statement holds for n and show that V(n) \implies V(n+1). We have that a_n < a_{n+1}. Adding 2 to both sides and taking square roots, we have that \sqrt{2+a_n} < \sqrt{2+a_{n+1}}, that is a_{n+1} < a_{n+2} by definition.
- Find bounds for a_n. The statement W(n): 0 < a_n < 2 holds for n=1, that is 0 < \sqrt2 < 2. Let us assume the statement holds for n and show that W(n) \implies W(n+1). We have that 0 < a_n < 2. Adding two and taking square roots, we have that 0 < \sqrt2 < \sqrt{2+a_n} < \sqrt4 = 2.
- The limit \lim_{n\to\infty} a_n exists, because (a_n)_{n=1}^\infty is a bounded monotonic sequence. Let A = \lim_{n\to\infty} a_n.
- Therefore the limit \lim_{n \to\infty} a_{n+1} exists as well and \lim_{n \to\infty} a_{n+1} = A. (For (n_k)_{k=1}^\infty = (2,3,4, \dots), we have that (a_{n_k})_{k=1}^\infty is a subsequence of (a_n)_{n=1}^\infty, from which the statement follows.)
- We have that a_{n+1} = f(a_n). That means that A = \lim_{n\to\infty} a_n = \lim_{n \to\infty} {f(a_n)} = f(\lim_{n \to\infty} a_n) = f(A) = \sqrt{2 + A}. Solving the equation A = \sqrt{2 + A}, we get A = -1 \lor A = 2.
- Putting it all together, we get that A = 2, because the terms of the sequence are increasing and a_1 > 0.
Is my solution correct?
Answer
Looks great. Here is a fun trick I've seen to answer this question.
Using the half angle formula, notice the following:
\cos\left(\frac{\pi}{4}\right)=\frac{1}{2}\sqrt 2\\\cos\left(\frac{\pi}{8}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt 2)}=\frac{1}{2}\sqrt{2+\sqrt 2}\\\cos\left(\frac{\pi}{16}\right)=\sqrt{\frac{1}{2}(1+\frac{1}{2}\sqrt{ 2+\sqrt2})}=\frac{1}{2}\sqrt{2+\sqrt {2+\sqrt 2}}\\\vdots\\\cos\left(\frac{\pi}{2^{n+1}}\right)=\underbrace{\frac{1}{2}\sqrt{2+\sqrt{2+\sqrt{2+{\ldots}}}}}_\text{n times}=\frac{1}{2}a_{n}
Now let n approach infinity.
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