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:




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


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.


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.