Sequence $a_{n+1}=\sqrt{k\cdot a_n}$

Question

I know that the sequence defined as $a_{n+1}=\sqrt{k+a_n}$ converges. But what about the sequence: $$ \begin{cases}a_1=\sqrt{k}\\a_{n+1}=\sqrt{k\cdot a_n}\end{cases}$$ where $k>1$? Does it converge?

I'd gladly hear any hints.

Answer 1

Hint: Define $b_n = \log_k(a_n)$. What does our recurrence become now?

Answer 2

A closed form is possible.

We have,

$$\ln a_{n+1}=\frac{1}{2}(\ln a_n+\ln k)$$

Let $b_n=\ln a_{n}$ then we have,

$$b_{n+1}=\frac{1}{2}b_{n}+\ln \sqrt{k}$$

Let $L$ be the real solution to,

$$L=\frac{1}{2}L+\ln \sqrt{k}$$

Then,

$$b_{n+1}-L=\frac{1}{2}(b_{n}-L)$$

So that,

$$b_{n}-L=(\frac{1}{2})^{n-1}(b_1-L)$$

$$b_{n}=(\frac{1}{2})^{n-1}(b_1-L)+L$$

$$a_{n}=e^{L}e^{(\frac{1}{2})^{n-1}(\ln a_1-L)}$$

$$a_{n}=e^{2\ln \sqrt{k}}e^{-(\frac{1}{2})^{n-1}\ln \sqrt{k}}$$

This shows that,

$$a_{n} \to e^{2 \ln \sqrt k}=k$$

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Actually,

$$a_1=\sqrt{k}=k^{\frac{1}{2}}$$

$$a_2=\sqrt{k \sqrt{k}}=k^{\frac{1}{2}}k^{\frac{1}{4}}$$

And that gives another way to see convergence to $k$ because,

$$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots=1$$

Answer 3

Yes, it does.

First, one can show by induction that $a_n \leqslant k$. Indeed, for $a_1$ it is true. If it is true for $a_m$ then it is also true for $a_{m+1}$:

$$a_{m+1} = \sqrt{ka_m} \leqslant \sqrt{k^2} = k$$

Then, you can show that the sequence is monotonically increasing. This is because

$$a_{n}(a_n - k) \leqslant 0 \Rightarrow a_n^2 \leqslant ka_n = a_{n+1}^2 \Rightarrow a_n \leqslant a_{n+1}$$

So, the sequence is increasing and bounded, which means that it has a limit.

We can also find this limit. Suppose that it equals to $L$. Then approaching $n \to \infty$ in the initial equation we obtain:

$$L = \sqrt{kL} \Rightarrow L = k$$

So, the limit of this sequence exists and equals to $k$.