1. ## Limit (1)

Suppose $\displaystyle x_1 > 0$ is given and let $\displaystyle x_{n+1}=\ln(1 + x_n), \ \ n \geq 1.$ Evaluate $\displaystyle \lim_{n\to\infty} nx_n.$

Spoiler:
The answer does not depend on the value of $\displaystyle x_1.$

2. Let $\displaystyle \lim_{n\to \infty}x_{n}=x$

$\displaystyle \lim_{n\to \infty}x_{n+1}=\lim_{n\to \infty}(ln(1+x_{n}))$

$\displaystyle x=ln(1+x)$

which is true only for $\displaystyle x=0$.

But $\displaystyle \lim_{n\to \infty}nx_{n}$ is something like

3. 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.

4. I also think that answer should be zero,but a rigorous solution must be there.

5. it's too soon to give away any part of solution. so i'll just give you a probably useless hint!

Spoiler:

Spoiler:
2

6. 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=$$\displaystyle lim_{n\rightarrow \infty } \frac{1}{\frac{(x'_{n-1})}{(x_{n-1})(ln(x_{n-1})^2)}}=lim_{n\rightarrow \infty }\frac{(x_{n-1})(ln(x_{n-1})^2)}{(x'_{n-1})}=lop..lop...lop..lop=x_1 I think my solution wrong right 7. Originally Posted by Amer since [/tex] use lopitals \displaystyle lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=lop=$$\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
How do you know that $\displaystyle \lim_{n\to \infty}\frac{1}{x_{n}}=\infty$

Therefore using L'Hopital rule is not proper

8. Originally Posted by pankaj
How do you know that $\displaystyle \lim_{n\to \infty}\frac{1}{x_{n}}=\infty$

Therefore using L'Hopital rule is not proper
since it is decreasing sequence

$\displaystyle x_{n+1}<x_{n}<x_{n-1}<.......<x_1$

$\displaystyle x_{n+1}=ln(x_{n})<x_{n}$

I am not sure about how I derive it ..

9. even if it's decreasing,we can't say that $\displaystyle \lim_{n\rightarrow\infty} \frac{1}{x_{n}} = \infty$
(i'm not sure,i'm just trying to help)

10. Originally Posted by Amer
since it is decreasing sequence

$\displaystyle x_{n+1}<x_{n}<x_{n-1}<.......<x_1$

$\displaystyle x_{n+1}=ln(x_{n})<x_{n}$

I am not sure about how I derive it ..
Caution: $\displaystyle x_{n+1}=\ln (1+x_n)$.

11. Originally Posted by arbolis
Caution: $\displaystyle x_{n+1}=\ln (1+x_n)$.
It still the same

$\displaystyle x_n>ln(x_{n}+1)$

the sequence is decreasing as I said try it

12. 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 $\displaystyle \lim_{n\to\infty} \frac{y_1+y_2+ \cdots + y_n}{n}=y.$

13. hi
$\displaystyle x_{n}\rightarrow 0$ while $\displaystyle n \rightarrow 0$
(thanks to "arbolis" )
then
$\displaystyle \lim_{n\rightarrow \infty}\left(\frac{1}{x_{n+1}}-\frac{1}{x_{n}} \right)= \frac{1}{2}$

14. Originally Posted by NonCommAlg

2) it is a well-known result that for any sequence $\displaystyle \{y_n \}$ if $\displaystyle \lim_{n\to\infty} y_n =y,$ then $\displaystyle \lim_{n\to\infty} \frac{y_1+y_2+ \cdots + y_n}{n}=y.$
...called Cesaro's mean, for those who're interested

15. Part 1: $\displaystyle \mathop {\lim }\limits_{n \to + \infty } x_n = 0$

Spoiler:
The sequence is decreasing and positive, hence it converges, let $\displaystyle L$ be its limit then: $\displaystyle \log \left( {1 + L} \right) = L$ and $\displaystyle L\geq{0}$ (since our sequence is positive), but this can happen only when $\displaystyle L=0$ ( See that $\displaystyle f\left( x \right) = x - \log \left( {1 + x} \right)$ is increasing in $\displaystyle \left( {0, + \infty } \right)$ )

Part 2: $\displaystyle \mathop {\lim }\limits_{n \to + \infty } \tfrac{{x_{n + 1} }} {{x_n }} = 1$

Spoiler:
$\displaystyle \tfrac{{x_{n + 1} }} {{x_n }} = \tfrac{{\log \left( {1 + x_n } \right)}} {{x_n }} = \tfrac{{x_n + o\left( {x_n } \right)}} {{x_n }} = 1 + o\left( 1 \right)$ as $\displaystyle n\to +\infty$ because $\displaystyle x_n\to {0^+}$

Part 3: $\displaystyle \left( {\tfrac{1} {{x_{n + 1} }} - \tfrac{1} {{x_n }}} \right) \to \tfrac{1} {2}$

Spoiler:
By Taylor (our sequence converges to 0): $\displaystyle \left( {\tfrac{1} {{x_{n + 1} }} - \tfrac{1} {{x_n }}} \right) = - \tfrac{{\log \left( {1 + x_n } \right) - x_n }} {{x_n x_{n + 1} }} = - \left( {\frac{{ - \tfrac{{x_n ^2 }} {2} + o\left( {x_n ^2 } \right)}} {{x_n x_{n + 1} }}} \right) = \tfrac{{x_n }} {{2x_{n + 1} }} + o\left( {\tfrac{{x_n }} {{x_{n + 1} }}} \right)$ as $\displaystyle n\to +\infty$ and we are done by Part 2

Part 4:
Computing $\displaystyle \lim_{n\to +\infty}{n\cdot x_n}$

Spoiler:
Since $\displaystyle \mathop {\lim }\limits_{n \to + \infty } \tfrac{{\left( {n + 1} \right) - n}} {{\tfrac{1} {{x_{n + 1} }} - \tfrac{1} {{x_n }}}} = 2$ by Part 3 , $\displaystyle x_n$ decreases and $\displaystyle \mathop {\lim }\limits_{n \to + \infty } \tfrac{1} {{x_n }} = + \infty$, by the Stolz-Cesaro Theorem (*) we have that $\displaystyle 2 = \mathop {\lim }\limits_{n \to + \infty } \tfrac{n} {{\tfrac{1} {{x_n }}}} = \mathop {\lim }\limits_{n \to + \infty } n \cdot x_n$

(*) A discrete version of L'Hôpital

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