The sequence $(a_n)_{n=1}^\infty$ is given by recurrence relation:

  • $a_1=\sqrt2$,
  • $a_{n+1}=\sqrt{2+a_n}$.

Evaluate the limit $\lim_{n\to\infty} a_n$.


  • 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?

  • 1
    $\begingroup$ It looks good. Since you prove the convergence though, why not put the "solving for A" part last? Also, I'm sure this has been asked and answered many times here, but I can't find the questions... $\endgroup$ – The Chaz 2.0 Jul 22 '13 at 16:42
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    $\begingroup$ Maybe you could just evaluate the infinite quadratic surd $x=\sqrt{2+\sqrt{2+...}}$.Then, $x=\sqrt{2+x}$ $\endgroup$ – chubakueno Jul 22 '13 at 18:51
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    $\begingroup$ @TheChaz2.0 Good idea, I made it that way. $\endgroup$ – David Jul 22 '13 at 18:52
  • $\begingroup$ @chu: this is always suggested, and then there is always a comment to remind us the "the convergence of the sequence needs to be proven." $\endgroup$ – The Chaz 2.0 Jul 22 '13 at 19:22
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    $\begingroup$ Almost certainly a duplicate--well, never mind. $\endgroup$ – Did Jul 22 '13 at 19:40

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.

  • $\begingroup$ Hello! Thank you. I will mark your answer -- the trick you posted is very nice. It is easy to see how it works from your post. $\endgroup$ – David Jul 22 '13 at 18:49

A somewhat shorter solution: prove by induction that $$a_n=2\cos \left(2^{-1-n}\pi\right).\tag{1}$$ Then, obviously, $\displaystyle \lim_{n\rightarrow\infty}a_n=2$.

Added: To understand the origin of (1), substitute $a_n=2\cos\alpha_n$ into the recursion relation.


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