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Let be the sequence $a_{n+1}=\ln{(1+a_n)}, n\ge1, a_1=1$.

Show that $\lim_{n\to\infty}{a_n}=0$ and the series $\sum_{n=1}^{\infty}{a_n^2}$ converges.

My try:

I assumed that $a_n$ is monotonic (decreasing) and lower bounded by $0$ (intuitively). I denoted $L=\lim_{n\to\infty}{a_n}$, so $L=\ln(1+L)$ which gives $L=0$. If there exists mistakes, please tell me.

How do I prove that the series converges? $$\sum_{n=1}^{\infty}{a_n^2}=\sum_{n=1}^{\infty}{\ln^2(1+\ln^2(1+\dots+\ln^2(1+1)))}$$

Thank you!

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Upps - I think your final sum-expression is not correct: you cannot put the squares inside the logarithms - then this would contradict your definition of $a_n=\ln(1+a_{n-1})$ – Gottfried Helms Jan 31 '14 at 17:04

By simple induction we see that $a_n\ge0$ forall $n$ and since $$\log(1+x)\le x,\quad \forall x\ge0$$ then the sequence $(a_n)$ is decreasing and bounded below so it's convergent to $\ell$ with $$\ell=\log(1+\ell)$$ hence $\ell=0$.

Now we have $$a_{n+1}=\log(1+a_n)=a_n-\frac{a_n^2}{2}+o(a_n^2)$$ hence $$\sum_{k=n}^\infty a_{k+1}-a_k=-a_n\sim-\frac{1}{2}\sum_{k=n}^\infty a_k^2$$ so the series is convergent.

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Step I. We show first that $\lim_{n\to\infty}a_n=0$.

For every $x>0$, we have that $$ 0<\log(1+x)<x, $$ and thus
$$ a_1>a_2>\cdots>a_n>a_{n+1}>\cdots >0, $$ and thus $a_n$ converges, with $a_n\to x\ge 0$. But $a_{n+1}=\log(1+a_n)\to \log(1+x)$. Thus $\log(1+x)=x$, which implies that $x=0$. Hence $a_n\to 0$.

Step II. We show now that $\lim_{n\to\infty}n\,a_n=2$.

We have $$ \frac{1}{a_{n+1}}-\frac{1}{a_n}=\frac{a_n-a_{n+1}}{a_na_{n+1}}=\frac{a_n}{a_{n+1}}\cdot \frac{1-\frac{\log(1+a_n)}{a_n}}{a_{n}}\to \frac{1}{2}, $$ since standard methods (i.e., l'Hôpital's rule) provide that $$ \lim_{x\to 0}\frac{1-\frac{\log(1+x)}{x}}{x}=\frac{1}{2}, $$ and also $$ \lim_{n\to\infty} \frac{a_n}{a_{n+1}}=\lim_{n\to\infty} \frac{a_n}{\log (1+a_{n})} =\lim_{x\to 0}\frac{x}{\log(1+x)}=1. $$ Using Stolz–Cesàro theorem $$ \lim_{n\to\infty}\frac{1}{na_n}=\lim_{n\to\infty}\frac{\frac{1}{a_n}}{n}= \lim_{n\to\infty}\frac{\frac{1}{a_{n+1}}-\frac{1}{a_n}}{(n+1)-n}= \lim_{n\to\infty}\left(\frac{1}{a_{n+1}}-\frac{1}{a_n}\right)=\frac{1}{2}. $$ This implies that the sequence $\{na_n\}$ is bounded, say by $M>0$, and hence $$ a_n^2\le \frac{M^2}{n^2}, $$ which implies that $\sum_{n=1}^\infty a_n^2<\infty$.

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A standard inequality is $$ x\ne0\implies1+x\lt e^x\tag{1} $$ Thus, $$ x_{n+1}=\log(1+x_n)\lt x_n\tag{2} $$ Furthermore, $$ x_n\gt0\implies x_{n+1}=\log(1+x_n)\gt0\tag{3} $$ Thus, $x_n$ is a decreasing sequence bounded below, therefore $\lim\limits_{n\to\infty}x_n$ exists. By continuity, $$ \lim_{n\to\infty}x_n=\lim_{n\to\infty}x_{n+1}=\log\left(1+\lim_{n\to\infty}x_n\right)\tag{4} $$ The contrapositive of $(1)$ says $$ 1+x\ge e^x\implies x=0\tag{5} $$ $(4)$ and $(5)$ say that $$ \lim_{n\to\infty}x_n=0\tag{6} $$

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