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(a) Consider a sequence defined by $a_0=1$ and $a_{n+1}=(\sqrt2)^{a_n}$ . Prove that limit exists and find it.

(b) Show that the limit doesn't exist finitely if we replace $\sqrt2$ by $1.5$. What are the maximum and minimum numbers such that limit exists?

I could do the first question and the first part of the second question.

For (a), by induction, each term of the sequence lies in $(1,2)$ and is therefore monotonically increasing by the recursion function. Hence the limit exists, let it be $l$, it comes $2$.

For (b), the sequence is monotone unbounded above, and the limit is $\infty$ similarly.

How do I do the second part of (b)?

Simply Beautiful Art
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Martund
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  • Find out why your proof of b works and try to replace $1.5$ by a smaller number. – markvs Sep 17 '20 at 02:59
  • Hint: if you plot different exponentials, you'll see that there are two intersections of the curves $t$ and $x^t$ which converge to one tangent intersection before they stop intersecting all together (as you modulate the value of $x$). How could you find the exact value of $x$ where a tangent intersection happens? – Ninad Munshi Sep 17 '20 at 03:00
  • @NinadMunshi, I think that happens for $e^{\frac1e}$, since at that point $a^t=t$ and $a^t\log a=1$. Solving gives $a=e^{\frac1e}$ and $t=e$. – Martund Sep 17 '20 at 03:05
  • But that should be the maximum value, the infimum should be $1$?? – Martund Sep 17 '20 at 03:06
  • Whoever downvoted it, you should justify that in comments. – Martund Sep 17 '20 at 03:07
  • @NinadMunshi, why don't you write that as an answer? I'll accept that. – Martund Sep 17 '20 at 03:08
  • Please state source. – jiten Sep 17 '20 at 03:10

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From my comment, we want to find when a tangent intersection of the curves $t$ and $x^t$ will occur. This means we have to satisfy the following system of equations:

$$\begin{cases}t = x^t \\ 1 = \log x \cdot x^t\\ \end{cases}$$

with the added constraint that we are looking for the solution $x > 1$

Ninad Munshi
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  • I don't think we need the constraint. From second equation we get $\log x=1/t$. Taking log on both sides of the first equation and using that, gives us $\log t=1$. Subtituting back gives us $\log x=1/e$. Hence $x=e^{\frac1e}.$ – Martund Sep 17 '20 at 03:16
  • And what is the infimum of all such numbers $x$?? – Martund Sep 17 '20 at 03:16
  • @Martund I think you can graph different curves and gain an intuition as to what happens. The max is the harder one to figure out. – Ninad Munshi Sep 17 '20 at 03:17