Limit (1)

**NonCommAlg** · Jun 7th 2009, 12:41 PM

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

Spoiler:

**pankaj** · Jun 8th 2009, 03:39 AM

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

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

$x=ln(1+x)$

which is true only for $x=0$ .

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

**arbolis** · Jun 8th 2009, 05:30 AM

If I'm not wrong the indetermination arises because $x_n \to 0$ while $n \to \infty$ .
By intuition $nx_n$ tends to $0$ as $n$ 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 $\ln (x)$ for $x=1$ is $1$ ...?

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.

**pankaj** · Jun 8th 2009, 06:35 AM

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

**NonCommAlg** · Jun 8th 2009, 03:21 PM

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

Spoiler:

**Amer** · Jun 8th 2009, 07:32 PM

since

$ln(r)<r$ for all real numbers

then the sequence

$x_{n+1}=ln(x_n)$ decreasing so the limit of it zero

$lim_{n\rightarrow \infty } nx_n =lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=\frac{\infty}{\infty}$ use lopitals

$lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=lop=$ $lim_{n\rightarrow \infty } \frac{1}{\frac{(x'_{n-1})}{(x_{n-1})(ln(x_{n-1})^2)}}=lim_{n\rightarrow \infty }\frac{(x_{n-1})(ln(x_{n-1})^2)}{(x'_{n-1})}=lop..lop...lop..lop=x_1$

I think my solution wrong right

**pankaj** · Jun 9th 2009, 06:44 AM

Originally Posted by Amer

since
[/tex] use lopitals

$lim_{n\rightarrow \infty } \frac{n}{\frac{1}{x_n}}=lop=$ $lim_{n\rightarrow \infty } \frac{1}{\frac{(x'_{n-1})}{(x_{n-1})(ln(x_{n-1})^2)}}=lim_{n\rightarrow \infty }\frac{(x_{n-1})(ln(x_{n-1})^2)}{(x'_{n-1})}=lop..lop...lop..lop=x_1$

I think my solution wrong right

How do you know that $\lim_{n\to \infty}\frac{1}{x_{n}}=\infty$

Therefore using L'Hopital rule is not proper

**Amer** · Jun 9th 2009, 08:24 AM

Originally Posted by pankaj

How do you know that $\lim_{n\to \infty}\frac{1}{x_{n}}=\infty$

Therefore using L'Hopital rule is not proper

since it is decreasing sequence

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

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

I am not sure about how I derive it ..

**Raoh** · Jun 9th 2009, 10:15 AM

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

)

**arbolis** · Jun 9th 2009, 01:50 PM

Originally Posted by Amer

since it is decreasing sequence

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

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

I am not sure about how I derive it ..

Caution: $x_{n+1}=\ln (1+x_n)$ .

**Amer** · Jun 10th 2009, 12:32 AM

Originally Posted by arbolis

Caution: $x_{n+1}=\ln (1+x_n)$ .

It still the same

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

the sequence is decreasing as I said try it

**NonCommAlg** · Jun 10th 2009, 03:58 AM

here are two important pieces of the solution. see if you can put everything together and give a complete solution:

1) find $\lim_{n\to\infty} \left(\frac{1}{x_{n+1}}-\frac{1}{x_n} \right).$

2) it is a well-known result that for any sequence $\{y_n \}$ if $\lim_{n\to\infty} y_n =y,$ then $\lim_{n\to\infty} \frac{y_1+y_2+ \cdots + y_n}{n}=y.$

**Raoh** · Jun 10th 2009, 04:26 AM

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

**Moo** · Jun 10th 2009, 10:33 AM

Originally Posted by NonCommAlg

2) it is a well-known result that for any sequence $\{y_n \}$ if $\lim_{n\to\infty} y_n =y,$ then $\lim_{n\to\infty} \frac{y_1+y_2+ \cdots + y_n}{n}=y.$

...called Cesaro's mean, for those who're interested