# Logarithm iteration convergence

I stumbled upon this while playing around with a calculator.

Take a number $$a_0$$ where $$1 \lt a_0 \lt 10^{10}$$ for reasons of dividing by 0 at some point.

So take that value and do this for the next number:

$$a_1=\dfrac{n}{\log_n{a_0}}$$

$$a_2=\dfrac{n}{\log_n{a_1}}$$

$$a_k=\dfrac{n}{\log_n{a_{k-1}}}$$

So somehow as $$k \to \infty \quad a_k \to n$$. This is of course for an arbitrary base $$n \in \mathbb{R}_1$$

How is this possible. And can one get the exact value of $$\displaystyle\lim_{k \to \infty} a_k$$ when that numerator is not necessarily n cause I tried the thing with numbers like 5 and 7 and the resulting limiting value looked irrational.

What I managed to do was to find out that $$\displaystyle\lim_{k \to \infty}a_k=\dfrac{n}{1-\log_n({1-\log_n({1-\log_n({1-\log_n({1-\log_n({ \ldots {1-(\log_n({\log_n({a_0})})})})})})})})}$$

The denominator in turn solves the functional equation $$\displaystyle1-\log_n{f(x)}=f(x)$$. I'm stuck there.

I hope I haven't overdone the Latex in the question

## 1 Answer

If the sequence converges, it must converge to a value such that

$$a=\frac{n}{\log_n a}$$ and this can be written

$$a\ln a=n\ln n.$$

For $$n>1$$, this equation has a single solution, $$a=n$$.

You can establish convergence by noting that the image of the interval $$[n-\delta,n+\delta]$$ is $$\left[\dfrac{n\ln n}{\ln(n+\delta)},\dfrac{n\ln n}{\ln (n-\delta)}\right]\approx\left[\dfrac{n\ln n}{\ln n+\dfrac\delta n},\dfrac{n\ln n}{\ln n-\dfrac\delta n}\right]\approx\left[n-\dfrac\delta{\ln n},n+\dfrac\delta{\ln n}\right]$$, which is a subset, provided $$n>e$$. At every step, the interval length shrinks by a factor $$\ln n$$, and convergence is linear.