Sequence $a_n$ is defined as $a_n = \ln (1 + a_{n-1})$, where $a_0 > -1$. Find all $a_0$ for which the sequence converges.
Using arithmetic properties of limits, I've found that if the sequence converges, then it converges to $0$, but I don't know how to prove it and find $a_0$.
$\endgroup$4 Answers
$\begingroup$You are correct in finding the limit in case of existence to be $0$
Notice that you are finding the fixed point of the function $$f(x)=\ln (1+x)$$
The derivative of your function is $$f'(x) = \frac {1}{1+x}$$
As you notice the derivative is positive for the domain of your function which is $x>-1$
We have $\frac {1}{1+x} <1$ for $x>0$ and $\frac {1}{1+x} >1$ for $-1<x<0$
Thuerefore the sequence will converge to $0$ for $x_0>0$ and diverge for $x_0<0$
$\endgroup$ 1 $\begingroup$hint
for $x>-1$$$\ln(1+x)\le x$$the sequence is decreasing.
if $-1<a_0<0$ the sequence is not well defined
if $a_0=0$ it is constant.
if $a_0>0$ the sequence is decreasing and positive.
the unique fixed point is zero . the sequence goes to zero.
$\endgroup$ $\begingroup$Note that we must have $a_n > 1$ for all $n$, else $a_{n+1}$ is not even well defined (I'll assume this silently throughout).
We note that $a_n \geq a_{n+1}$ if and only if $e^{a_n}\geq 1 + a_n$. Given the power series expansion of $e^x$, we have $$e^{a_n} - (1 + a_n) = \sum\limits_{i=2}^{\infty}\frac{a_n^i}{i!},$$ which is clearly non-negative for all $a_n$. Thus, the sequence is decreasing, and hence monotonic, for any $a_0$.
Now, decreasing sequences converge if and only if they are bounded below, and given what you already know about what it must converge to, you essentially already know the answer, so we just need to show that the sequence is bounded below for $a_0 \geq 0$, but not for $a_0 < 0$.
We'll do those in order: if some $a_n \geq 0$, then $a_{n+1} = \ln(1 + a_n) \geq \ln(1) = 0$, so indeed the sequence is bounded below by $0$ for $a_0 \geq 0$, and so converges for all such values.
On the other hand, if $a_0 < 0$, then, since our sequence is decreasing, it cannot converge to $0$ (it ever gets within a distance $|a_0|$ of $0$), so, by the argument that you already have, cannot converge.
Thus, our sequence converges exactly for $a_0 \geq 0$.
$\endgroup$ $\begingroup$HINT
We can show that as $a_0 >0$
- $a_n$ is decreasing
- $a_n$ is bounded below
