An issue with approximations of a recurrence sequence

By trying to give an approximation to a given recurrence sequence I encountered a problem.

To be more precise I have a method but it fails if the right condition is not met and I wonder how I should proceed then.

An example will clarity.

General Question:

Let $a_0=c>0$ , $a_n=f(a_{n-1})$ and the sequence is strictly decreasing and has limit equal to $0$. Also for every finite $n$ we have $a_n >0$.

Specific Example:

Let $f$ be the function $log(1+x)$.

THE NECC CONDITION : $\dfrac{x^m}{f(x)}$ is analytic at $x=0$ for some integer $m$.

Solution for the Specific Example:

Step 1. Since $$0 < a_{n+1} = \log (1+a_n) < a_n,$$ it is a monotone decreasing sequence which is bounded. Thus it must converge to some limit, say $\alpha$. Then $\alpha = \log(1 + \alpha)$, which is true precisely when $\alpha = 0$. Therefore it follows that $$a_n = o(1). \tag{1}$$

Before going to the next step, we make a simple observation: it is easy to observe that the function $$\frac{x}{\log(1+x)}$$ is of class $C^3$. In particular, whenever $|x| \leq \frac{1}{2}$, we have $$ \frac{x}{\log (1+x)} = 1+\frac{x}{2}-\frac{x^2}{12}+O(x^3). $$ This can be rephrased as $$ \frac{1}{\log(1+x)} = \frac{1}{x}+\frac{1}{2}-\frac{x}{12}+O(x^2). \tag{2}$$ Here, we note that the bound, say $C > 0$, for the Big-Oh notation does not depend on $x$ whenever $|x| \leq \frac{1}{2}$.

Step 2. By noting $(1)$, we fix a positive integer $N$ such that whenever $n \geq N$, we have $|a_n| \leq \frac{1}{2}$. Then by $(2)$, $$ \frac{1}{a_{n+1}} - \frac{1}{a_n} = \frac{1}{2} + O(a_n), $$ where the bound for Big-Oh notation depends only on $N$. Indeed, we may explicitly choose a bounding constant as $$C'=\frac{1}{12} + \frac{1}{2}C,$$ where $C$ is the same as in $(2)$. Thus if $n > m > N$, we then have $$ \begin{align*} \frac{1}{a_n} &= \frac{1}{a_{m}} + \sum_{k=m}^{n-1} \left( \frac{1}{a_{k+1}} - \frac{1}{a_k} \right) \\ &= \frac{1}{a_{m}} + \sum_{k=m}^{n-1} \left( \frac{1}{2} + O(a_k) \right) \\ &= \frac{1}{a_{m}} + \frac{n-m}{2} + O((n-m)a_m). \end{align*} $$ Thus we have $$ \left|\frac{1}{n a_n} - \frac{1}{2}\right| \leq \frac{1}{n}\left(\frac{1}{a_m} + \frac{m}{2} + C'm a_m \right) + C' a_m.$$ Taking limsup as $n\to\infty$, we have $$ \limsup_{n\to\infty}\left|\frac{1}{n a_n} - \frac{1}{2}\right| \leq C' a_m. $$ Since now $m$ is arbitrary, the right-hand side can be made as small as we wish. Thus the left-hand side must vanish, yielding $$ \frac{1}{n a_n} = \frac{1}{2} + o(1),$$ or equivalently $$ n a_n = 2 + o(1). \tag{3} $$

However notice how much that method depended on the NECC CONDITION , SINCE we needed a taylor series with nonzero radius when expanded from $0$.

So my question is what should I do if that NECC condition is not met for $f$ ?

MORE SPECIFIC How do I proceed for an $f(x)$ that is not meromorphic at $0$ but is meromorphic on the halfopen interval $)0,1/2)$ ?

I considered using the taylor expansion of $f$ at $\frac{1}{q}$ and taking the limit for large positive $q$ going to $\infty$. But Im not sure how to do that. Or even if its a good idea.

Im also unsure if it matters what type of singularty we have at $0$.

But I believe there must be a way!

ONE MORE THING

I assumed that $f$ has no natural boundary. Not sure if it matters or not. Such as if $f(x)$ is defined for $x<0$ matters or not.

To end this question maybe a more specific question : ADD to the above question the condtion that $f(f(x)-x) - (f(x)-x) = g(x)$ where $g$ is the FUNCTIONAL INVERSE of an entire function.

Maybe that makes the question easier because it is more specific.

Solution 1:

Hence the question [is] how to handle it.

"It" corresponds to $a_0=c$ with $c\gt0$ and $a_{n+1}=f(a_n)$ for every $n\geqslant0$, where $$ f(x)=\frac1{\frac1x+\log\frac1x}. $$ Such a function cries out for conjugacy... Thus, let us consider the sequence defined by $b_n=1/a_n$, then $b_{n+1}=g(b_n)$ for every $n\geqslant0$, where $$ g(x)=x+\log x. $$ If $b_0\gt1$, that is, if $c\lt1$, then $(b_n)$ is increasing and $b_n\to+\infty$ hence $a_n\to0$. If $b_0=1$, that is, if $c=1$, then $b_n=a_n=1$ for every $n$. If $b_0\lt1$, that is, if $c\gt1$, then $b_n\lt0$ after a while hence $(b_n)$ is not defined.

Thus, the case of interest is $c\lt1$. Associated to $b_{n+1}=g(b_n)$ is the ordinary differential equation $x'(t)=g(x(t))-x(t)=\log x(t)$, solved by $$ t=\int_{x(0)}^{x(t)}\frac{\mathrm du}{\log u}=\mathrm{li}(x(t))-\mathrm{li}(x(0)), $$ where $\mathrm{li}$ denotes the logarithmic integral. Thus, $\mathrm{li}(x)\sim x/\log x$ and $x(t)\sim t\log t$ when $t\to+\infty$. This suggests, and indeed one can show, that $b_n\sim n\log n$ when $n\to\infty$.

More precisely, $\log t\geqslant\log b_n$ for every $t$ in $(b_n,b_{n+1})$ hence $$ \mathrm{li}(b_{n+1})-\mathrm{li}(b_n)=\int_{b_n}^{b_{n+1}}\frac{\mathrm dt}{\log t}\leqslant\frac{b_{n+1}-b_n}{\log b_n}=1, $$ that is, $$ \mathrm{li}(b_n)\leqslant\mathrm{li}(b_0)+n. $$ Likewise, $b_{n+1}-\log b_{n+1}\leqslant b_n$ hence $\log(t-\log t)\leqslant\log b_n$ for every $t$ in $(b_n,b_{n+1})$ hence $$ \int_{b_n}^{b_{n+1}}\frac{\mathrm dt}{\log( t-\log t)}\geqslant\frac{b_{n+1}-b_n}{\log b_n}=1, $$ that is, $$ \int_{b_0}^{b_{n}}\frac{\mathrm dt}{\log( t-\log t)}\geqslant n. $$ Since $\log( t-\log t)\sim\log t$ when $t\to\infty$, all this yields $\mathrm{li}(b_n)\sim n\sim b_n/\log b_n$, and finally, $$ \lim_{n\to\infty}n(\log n)a_n=1. $$