Find a formula for a sequence $\{\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}},...\}$

Question

I'm trying to find a formula for the following sequence:

$\{\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}},...\}$

I thought of solving it recursively and I got this formula:

$a_{n}=\sqrt{3*a_{n-1}}$

$a_{0}=1$

Is there a better and non-recursive formula for the given sequence?

I don't have a solution at first glance but I will comment that there is no need to designate 1 as the 0th element. You could just start with a_1 = sqrt(3) and the recursion. — change_picture, Jun 25 '16 at 20:47
The sequence can be written as ${ 3^{1/2}, 3^{1/2+1/4}, 3^{1/2+1/4+1/8}, \dots}$, so $a_n = 3^{\sum_{k=1}^n 1/2^k} = 3^{1-2^{-n}}$. — user217285, Jun 25 '16 at 20:48
You can consider $a_n$ as $3^{1/2}\cdot 3^{1/4} \cdot ... \cdot 3^{1/2^n}$ and then use the formula $\sum_{i=1}^n 1/2^i = 1-\frac{1}{2^{n}}$. — Levent, Jun 25 '16 at 20:48
See also: $\sqrt{7\sqrt{7\sqrt{7\sqrt{7\sqrt{7\cdots}}}}}$ approximation. — Martin Sleziak, Sep 12 '17 at 04:29

score 16 · Accepted Answer · answered Jun 25 '16 at 20:46

16

If we start with $a_0$, what about $a_n=3^{\left(1-\frac{1}{2^{n+1}}\right)}$? Note that the terms are $3^{1/2}$, $3^{3/4}$, $3^{7/8}$, and so on.

answered Jun 25 '16 at 20:46

André Nicolas

514,336

score 2 · Answer 2 · answered Jun 25 '16 at 21:56

Let $b_n = \ln( a_n )$, then according to $a_n = \sqrt{3a_{n-1}}$, we have:

$b_n = \frac{1}{2}[ \ln(3)+b_{n-1} ]$ with $b_0=0$

which is a classical problem. We can easily find its solution: $b_n = \ln(3)[ 1 - (\frac{1}{2})^n]$. It is trivial to convert $b_n$ to $a_n = 3^{1-(\frac{1}{2})^n}$

Find a formula for a sequence $\{\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}},...\}$

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