let $a_{n+1} = \log (1+a_n) $ for $n \ge 1$ . Check whether $\sum_{n=1}^{\infty} a_n$ converge or diverge?

Question

let $a_1$ be an arbitrary postive number and let $a_{n+1} = \log (1+a_n) $ for $n \ge 1$. Check whether $\sum_{n=1}^{\infty} a_n$ converges or diverges?

My attempt: I know that $\log(1+x) = x - \frac{x^2}{2} + \frac {x^3}{3} \cdots$

Now here how can I conlcude that $\sum_{n=1}^{\infty} a_n$ converges or diverges?

Any hints/solution will be appreciated.

Answer 1

Look for hints within math.SE! Argue that $$\lim_{n\to\infty} n\cdot a_n=2$$ by referring to $a_{n+1}=\log(1+a_n),~a_1>0$. Then find $\lim_{n \rightarrow \infty} n \cdot a_n$. Therefore the series $\sum a_n$ diverges by comparison to the harmonic series.

Answer 2

Just a hint

Find $\beta$ such that $a_{n+1}^\beta - a_n^\beta$ converges. Based on that you can find an equivalent to $a_n$ and determine if $\sum a_n$ converges or not.

Answer 3

At least one quickly sees that $a_n$ is strictly decreasing towards $0$, which is a necessary condition for convergence of the series.

However, the convergence of the $a_n$ is not fast enough: If $\frac1{2N}<a_n<\frac1N$, then $a_{n+1}>a_n\cdot\left(1-\frac{a_n}2\right)>a_n\cdot\left(1-\frac1{2N}\right)$. By induction, $a_{n+k}>a_n\cdot\left(1-\frac1{2N}\right)^k$, and by Bernoulli's inequality, $a_{n+k}>a_n\cdot\left(1-\frac k{2N}\right)$ . In particular, $a_{n+k}>\frac12 a_n$ for $k<N$, hence the contribution of $N$ terms starting at $a_n$ is at least $N\frac12a_n>\frac14$.

We conclude that $\sum a_n$ diverges.