Suppose $\displaystyle x_1 > 0$ is given and let $\displaystyle x_{n+1}=\ln(1 + x_n), \ \ n \geq 1.$ Evaluate $\displaystyle \lim_{n\to\infty} nx_n.$
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If I'm not wrong the indetermination arises because $\displaystyle x_n \to 0$ while $\displaystyle n \to \infty$.
By intuition $\displaystyle nx_n$ tends to $\displaystyle 0$ as $\displaystyle n$ tends to positive infinite. However I'm not sure at all, and I guess it could tend to a constant. This is a tough problem!
Or is it 1 because the slope of $\displaystyle \ln (x)$ for $\displaystyle x=1$ is $\displaystyle 1$...?
Edit : Now my question is... can we prove it with Peano's axioms? Or we have to suppose another set of axioms?
I'm sad I don't have enough time to dedicate myself fully to the problem.
since
$\displaystyle ln(r)<r $ for all real numbers
then the sequence
$\displaystyle x_{n+1}=ln(x_n) $ decreasing so the limit of it zero
$\displaystyle lim_{n\rightarrow \infty } nx_n =lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=\frac{\infty}{\infty}$ use lopitals
$\displaystyle lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=lop=$$\displaystyle lim_{n\rightarrow \infty } \frac{1}{\frac{(x'_{n-1})}{(x_{n-1})(ln(x_{n-1})^2)}}=lim_{n\rightarrow \infty }\frac{(x_{n-1})(ln(x_{n-1})^2)}{(x'_{n-1})}=lop..lop...lop..lop=x_1$
I think my solution wrong right
here are two important pieces of the solution. see if you can put everything together and give a complete solution:
1) find $\displaystyle \lim_{n\to\infty} \left(\frac{1}{x_{n+1}}-\frac{1}{x_n} \right).$
2) it is a well-known result that for any sequence $\displaystyle \{y_n \}$ if $\displaystyle \lim_{n\to\infty} y_n =y,$ then $\displaystyle \lim_{n\to\infty} \frac{y_1+y_2+ \cdots + y_n}{n}=y.$
Part 1: $\displaystyle
\mathop {\lim }\limits_{n \to + \infty } x_n = 0
$
Spoiler:
Part 2: $\displaystyle
\mathop {\lim }\limits_{n \to + \infty } \tfrac{{x_{n + 1} }}
{{x_n }} = 1
$
Spoiler:
Part 3: $\displaystyle
\left( {\tfrac{1}
{{x_{n + 1} }} - \tfrac{1}
{{x_n }}} \right) \to \tfrac{1}
{2}
$
Spoiler:
Part 4: Computing $\displaystyle \lim_{n\to +\infty}{n\cdot x_n}$
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