How to check the convergence or divergence of sequence $\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{\sqrt{3}\sqrt{3}},\dots$?

Question

I tried like this

$a_n=\sqrt{3\sqrt{3}\sqrt{3}\dots}$ $n-times$

Taking square of the both sides

$(a_n)^2=3a_{n-1}$

$\Rightarrow \dfrac{(a_n)^2}{a_{n-1}}=3$

But I don't know how to proceed further.

Answer 1

Note that $a_n$ can also be expressed in terms of exponent notation. $$a_n=\sqrt{3}\times\sqrt{\sqrt{3}}\times\ldots\times\sqrt{\sqrt{\ldots\sqrt{3}}}\\ =3^\frac{1}{2}\times3^\frac{1}{4}\times\ldots\times3^\frac{1}{2^n}\\ =3^{\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^n}}\\ \stackrel{\text{geometric sum}}{=}3^{1-\frac{1}{2^n}}$$

$$\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}3^{1-\frac{1}{2^n}}\\ =3^{1-\lim_{n\rightarrow\infty}\frac{1}{2^n}}\\ =3^{1-0}=3$$

This shows that $a_n$ converges to 3.