Vtyjkyuk

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 $log(1+x) = x - fracx^22 + frac x^33 cdots$

Now here how can I conlcude that $sum_n=1^infty a_n$ converges or diverges?

Any hints/solution will be appreciated.

Look for hints within math.SE! Argue that $$lim_ntoinfty ncdot a_n=2$$ 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.

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 $frac12N<a_n<frac1N$, then $a_n+1>a_ncdotleft(1-fraca_n2right)>a_ncdotleft(1-frac12Nright)$. By induction, $a_n+k>a_ncdotleft(1-frac12Nright)^k$, and by Bernoulli's inequality, $a_n+k>a_ncdotleft(1-frac k2Nright)$ . In particular, $a_n+k>frac12 a_n$ for $k<N$, hence the contribution of $N$ terms starting at $a_n$ is at least $Nfrac12a_n>frac14$.

We conclude that $sum a_n$ diverges.

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.