Find $\lim_{n \to \infty } a_n$

Question

\begin{cases} a_1=\sqrt 3 \\ a_2 = \sqrt {3\sqrt 3}\\ a_n = \sqrt {3a_{n-1}} \quad \text{for } n\in\mathbb Z^+\end{cases}

This sequence is bounded above by $3$ and is monotone increasing, so by monotone bounded sequence theorem, the sequence converges.

But, the question asks to find $\lim_\limits{n \to \infty } a_n$. I guess the limit is $3$, but don't know how to prove it.

Could you give some hint? Thank you in advance.

Answer 1

$$a_n=3^{1/2+1/4+\cdots+1/2^n}$$

Now $S(n)=\dfrac12+\dfrac14+\cdots+\dfrac1{2^n}=\dfrac12\left(\dfrac{1-\left(\dfrac12\right)^n}{1-\dfrac12}\right)$

$\lim_{n\to\infty}S(n)=\dfrac12\left(\dfrac{1-0}{1-\dfrac12}\right)=1$

Answer 2

Hint

If $\ell$ is the limit, then $$\ell=\sqrt{3\ell}.$$

Answer 3

Suppose $a_n\to l$ for $n\to \infty$. Then also $a_{n+1}\to l$ because $n+1\sim n$ as $n\to\infty$. So $$\lim a_n=\lim a_{n+1}=\lim3\sqrt{a_n}\implies l=3\sqrt{l}$$

Answer 4

Suppose that $a$ is the limit. Then:

$$a = \sqrt{3\sqrt{3\sqrt{3\sqrt{\ldots}}}}.$$

You can write that:

$$a = \sqrt{3a}.$$

This means that:

$$a = \sqrt{3}\sqrt{a} \Rightarrow \sqrt{a} = \sqrt{3} \Rightarrow a = 3.$$

Anyway, we don't know if this sequence will converge to the limit $a$.

To this aim, notice that:

$$\frac{\partial a_{n}}{\partial a_{n-1}} = \frac{1}{2}\sqrt{\frac{3}{a_{n-1}}}.$$

For $a_{n-1} = a = 3$, we get that the value of this derivative is $\frac{1}{2}$ which is in modulus less than $1$. Then, the sequence converge to $a= 3$.

Answer 5

Set $b_n := \ln a_n$:

• $b_1 = \frac{1}{2}\ln 3$
• $b_{n} = \frac{1}{2}\ln{(3a_{n-1})} = \frac{1}{2}\ln 3 + \frac{1}{2}\ln a_{n-1} = \frac{1}{2}\ln 3 + \frac{1}{2}b_{n-1}$
• $\Rightarrow b_n = \left(\frac{1}{2}+\cdots + \frac{1}{2^n} \right)\ln 3= \left(1- \frac{1}{2^n} \right)\ln 3 \stackrel{n \rightarrow \infty}{\rightarrow}\ln 3$
• $a_n = e^{b_n} \stackrel{n \rightarrow \infty}{\rightarrow} e^{\ln 3} = 3$