$\{\log n-n(1-na_n)\}$ is convergent for $a_{n+1}=a_n(1-a_n)$ and $0

Question

Let a sequence $\{a_n\}_{n=1}^\infty$ satisfy $0<a_1<1$ and $a_{n+1}=a_n(1-a_n)$ for $n\geq 1$. Prove that $\{\log n-n(1-na_n)\}$ is convergent (without using asymptotic analysis).

This problem came to me when I was preparing my exercise lessons on Analysis(I) course, as a teaching assistant. The original problem asks students to prove $\lim_{n\to\infty}na_n=1$. I'm curious about the asymptotic behaviour of this sequence. Using the method of asymptotic analysis, I found that $$a_n= \frac1n-\frac{\log n}{n^2}+\frac C{n^2}+o\left(\frac1{n^2}\right),\qquad n\to\infty,$$ where $C$ is a constant probably depending on the value of $a_1$. After that, I tried to prove this asymptotic relation, by directly proving the following three properties:

• $\displaystyle\lim_{n\to\infty}na_n=1$.
• $\displaystyle\lim_{n\to\infty}\frac{n}{\log n}(1-na_n)=1$.
• $\displaystyle\lim_{n\to\infty}\log n\left(1-\frac{n}{\log n}(1-na_n)\right)=\lim_{n\to\infty}\left(\log n-n(1-na_n)\right)$ exists.

We can easily show that $a_n$ is decreasing to $0$. The first two properties can be proved using $\frac1{a_{n+1}}=\frac1{a_n}+\frac1{1-a_n}$ and Stolz–Cesàro theorem: for the first one, we apply Stolz–Cesàro theorem to $\frac{n}{1/a_n}$; and for the second one, we apply Stolz–Cesàro theorem to $$\frac{\frac{1-na_n}{a_n}}{\log n}=\frac{\frac1{a_n}-n}{\log n}.$$

However, I don't know how to prove the third one. I don't believe that we can prove it by directly using Stolz–Cesàro theorem. Because Stolz–Cesàro theorem will give a limit not depending on $a_1$.

Any help would be appreciated!

Answer 1

I don't get what "without asymptotic analysis" means.

So let's redo the solution (using Stolz–Cesàro theorem). Given $a_{n+1}=a_n(1-a_n)$ with $0<a_1<1$, we get $a_n\to 0$ as $n\to\infty$ easily. Next, we put $a_n=1/b_n$; then $b_n\to\infty$ and $$b_{n+1}-b_n=1+\frac1{b_n-1}\to1,$$ thus $b_n/n\to1$ by SCT. Now we put $b_n=n+c_n$; then $c_n=o(n)$ and $$c_{n+1}-c_n=\frac1{n+c_n-1}\implies\frac{c_{n+1}-c_n}{\log(n+1)-\log n}\to1,$$ and SCT gives $c_n/\log n\to1$. Finally put $c_n=\log n+d_n$; then $d_n=o(\log n)$ and $$d_{n+1}-d_n=\frac1{n+\log n+d_n-1}-\log\left(1+\frac1n\right)=O\left(\frac{\log n}{n^2}\right),$$ thus $\sum\limits_{n=1}^\infty(d_{n+1}-d_n)$ converges (absolutely) and $\lim\limits_{n\to\infty}d_n=d_\infty$ exists.

Plugging everything back, we have $a_n=1/(n+\log n+d_n)$, and $$\log n-n(1-na_n)=-d_n+\frac{(\log n+d_n)^2}{n+\log n+d_n}$$ tends to $-d_\infty$ as $n\to\infty$.