4
$\begingroup$

\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.

$\endgroup$
5
$\begingroup$

Hint

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

$\endgroup$
8
$\begingroup$

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

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

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

$\endgroup$
1
$\begingroup$

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 $$\lim a_n=\lim a_{n+1}=\lim3\sqrt{a_n}\implies l=3\sqrt{l}$$

$\endgroup$
0
$\begingroup$

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 $a= 3$.

$\endgroup$
  • $\begingroup$ Or $a=0$ of course, when you say $a = \sqrt{3}\sqrt{a}$, which in this particular case is not the solution. $\endgroup$ – Václav Mordvinov Apr 25 '18 at 10:15
  • $\begingroup$ @VáclavMordvinov $a=0$ is obviously unstable. $a_0 = \sqrt{3}$ belongs to the basin of attraction of $a=3$. This basin is $\mathbb{R}^+$. $\endgroup$ – the_candyman Apr 25 '18 at 10:19
0
$\begingroup$

Set $b_n := \ln a_n$:

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.