3

Prove that the sequence $\sqrt{2}, \sqrt{2+\sqrt{2}}, \sqrt{2+\sqrt{2+\sqrt{2}}},...$ converges and find its limit. Prove this without using derivatives (Source: Based on Abbot's Analysis).

Proof (partial)

Let $f$ be an increasing function with fixed point $k$, such that $x \leq k \implies x \leq f(x) \leq k$. Let $a$ be a sequence $a_1 \leq k, a_{n+1} = f(a_n)$. By induction, $a$ is bounded and increasing, and therefore converges. Furthermore, if $x < k \implies f(x) > x$, $a$ converges to $k$.

Let $g(x) = \sqrt{x+2}, g > 0$. $g$ is increasing, with fixed point $g(2) = 2$. By properties of ordered fields, $x < 2 \implies g(x) < 2$. It remains to show that $x < 2 \implies g(x) > x$. This can be proven by taking the derivative of $x - \sqrt{x+2}$, but I'm not able to prove it without derivatives.

$a_1 = \sqrt{2} < 2, a_{n+1} = g(a_n)$, QED.

Questions

  1. Can you help me fill in the missing part?
  2. Critique what I've done (both the proof and the writing).
  3. It seems that functions with the property $f(x) > x$ would be very important for studying series. Is there a name for them? Do they have any other salient properties?
  4. Instead of simply tackling the problem locally, in terms of $\sqrt{2}$, I abstracted it to more general series. Is that advisable? What are your thoughts on that?
SRobertJames
  • 4,278
  • 1
  • 11
  • 27
  • If you want to show $\sqrt{x+2}>x$ for $-2<x<2$, it's equivalent to show $x+2>x^2 \text{sign}(x)$ over the same range; why is this more tractable? – Integrand Aug 06 '21 at 15:48
  • 3
    You can compare your proof with the versions here and here. – rtybase Aug 06 '21 at 16:03
  • if $-1<x<2$ then $x^2-x-2=(x+1)(x-2)<0$ so $x^2<x+2$ so $x<g(x)$ – J. W. Tanner Aug 06 '21 at 16:39
  • In regard to your 4th question, you have overgeneralized. Consider the function $f(x) = \begin{cases} \frac 12x+1 &(x<2) \ \frac 12x+2 &(x\ge 2) \end{cases}$ with $k=4$ and $a_1=0$. This meets the conditions under which you claim $a_n$ converges to $k$, but in fact $a_n$ converges to $2$. – tuna Aug 06 '21 at 17:31
  • For a completely different approach to the problem, try applying the identity $x-y = \dfrac{x^2-y^2}{x+y}$ to $x=2$, $y=a_n$. (I still think you should pursue your original idea too, but I couldn't resist posting a cute alternative!) – tuna Aug 06 '21 at 17:44
  • Why not just solving the equation and pick the positive solution $(1+\sqrt{5})/2$? – Moti Aug 06 '21 at 19:02
  • @Moti: did you mean the equation $\sqrt{x+2}=x$, which has positive solution $2$? – J. W. Tanner Aug 06 '21 at 19:32
  • @Moti, the fixed point is $2$, not $(1+\sqrt 5)/2$. But more importantly, that doesn't prove convergence. Compare the sequence $a_1=1$, $a_{n+1}=2a_n+1$. The equation $x=2x+1$ has solution $x=-1$ but this is not the limit of $a_n$. – tuna Aug 06 '21 at 19:33
  • @Tuna Assuming convergence leads to the equation $x^2 = 2 +x$ – Moti Aug 07 '21 at 01:53
  • 1
    @J.W.Tanner yes. – Moti Aug 07 '21 at 01:56
  • @Tuna It seems you right and my assumption for the equation was not right. – Moti Aug 07 '21 at 02:04

0 Answers0