Find the limit of the sequence $x_{n+1}=\sqrt{3x_n}$

Question

Let $x_{n+1}=\sqrt{3x_n}$ and $x_1=1$. Prove $x_n=3^{1-(\frac{1}{2^{n-1}})}$ for all $n$ and find the limit of $\{x_n\}$.

Notes: The first few terms of the sequence are $1,\sqrt{3},\sqrt{3\sqrt{3}},\sqrt{3\sqrt{3\sqrt{3}}} ...$ I do not know from this how to find the equation for the $x_n$ term in terms of $3$. The second part can be answered as follows, please tell me if my thinking is correct. \begin{align*} x_n&=3^{1-(\frac{1}{2^{n-1}})} \\ \lim_{n\rightarrow \infty}x_n&=\lim_{n\rightarrow \infty}3^{1-(\frac{1}{2^{n-1}})} \\ &=3^{1-\lim_{n\rightarrow \infty}(\frac{1}{2^{n-1}})} \\ &=3^{1-(\frac{1}{\lim_{n\rightarrow \infty}2^{n-1}})} \\ &= 3 \end{align*} Also I have looked at Show that the sequence $\sqrt{2},\sqrt{2\sqrt{2}},\sqrt{2\sqrt{2\sqrt{2}}},...$ converges and find its limit. But there is no discussion on how to find something similar to $x_n=3^{1-(\frac{1}{2^{n-1}})}$ .

Answer 1

Your proof for the second part is correct.

What's nice here is you don't need to find the equation for $x_n$, it's already given to you. You just have to prove the equation is correct. You can do this by induction on $n$. The base case $n=1$ holds since $x_1=1=3^{1-\frac1{2^{1-1}}}$.

Assume that $x_n=3^{1-\frac1{2^{n-1}}}$. Then $$ x_{n+1}=\sqrt{3x_n}=\sqrt{3\cdot 3^{1-\frac1{2^{n-1}}}}=3^{1-\frac1{2^{(n+1)-1}}} $$ This proves the formula for $n+1$, completing the proof by induction.

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One last comment: in general, there is no procedure for getting a formula for $x_n$ given a recursive equation is solves. In this case, it could be done by looking at $x_n$ for several small values of $n$, simplifying as much as possible, then noticing a pattern.

Answer 2

If you want to see where the formula comes from, notice the pattern:

\begin{eqnarray} x_1 &=& 1\\ x_2 &=& \sqrt{3} &=& 3^{\frac12}\\ x_3 &=& \sqrt{3\sqrt{3}} &=& \bigl(3\cdot3^{\frac12}\bigr)^\frac12 &=& 3^\frac12 \cdot 3^\frac14 &=& 3^{\frac12 + \frac14}\\ x_4 &=& \sqrt{3\sqrt{3\sqrt{3}}} &=& \left(3\bigl(3\cdot3^{\frac12}\bigr)^\frac12\right)^\frac12 &=& 3^\frac12 \cdot 3^\frac14 \cdot 3^\frac18 &=& 3^{\frac12 + \frac14 + \frac18}\\ \end{eqnarray}

In general, the sequence can be written

$$x_n = 3^{\sum_{k=1}^{n-1}\frac{1}{2^k}}$$

Inductively,

\begin{eqnarray} x_{n+1} &=& \sqrt{3x_n}\\ &=& \sqrt{3 \cdot 3^{\sum_{k=1}^{n-1}\frac{1}{2^k}}}\\ &=& \sqrt{3^{1+\sum_{k=1}^{n-1}\frac{1}{2^k}}}\\ &=& 3^{\frac12+\frac12\sum_{k=1}^{n-1}\frac{1}{2^k}}\\ &=& 3^{\sum_{k=1}^{(n-1)+1}\frac{1}{2^k}}\\ \end{eqnarray}

The sum for the geometric series is $\displaystyle\sum_{k=0}^{n-1} a r^k = a \dfrac{1-r^n}{1-r}$, so

$$1+\sum\limits_{k=1}^{n-1}\frac{1}{2^k} = \sum\limits_{k=0}^{n-1}\frac{1}{2^k} = \frac{1-(\frac12)^n}{1-\frac12} = \frac{1-\frac{1}{2^n}}{\frac12} = 2 - \frac{1}{2^{n-1}}$$

Subtracting $1$ from both sides of the equation, we can write

$$x_n = 3^{\left(1 - \frac{1}{2^{n-1}}\right)}= 3^{1-2^{1-n}}$$

Answer 3

Hint

Considering the sequence $$x_{n+1}=\sqrt{3x_n}$$ take logarithms of both sides $$\log(x_{n+1})=\frac{\log(3)}2+\frac 12 \log(x_n)$$ Now define $y_n=\log(x_n)$ which makes the new sequence to be $$y_{n+1}=\frac 12 y_n+\frac{\log(3)}2$$ which is much easier to manipulate.

For the general sequence $$y_{n+1}=a y_n+b\qquad (y_1=0)$$ the general solution is simply $$y_n=\frac{1-a^{n-1}}{1-a} b$$ So, if $a<1$, the limit is just $\frac b {1-a}$; then the limit for the original sequence.