1. recursive sequence

hi,
i need to find the limit of the following sequence
lim (an)1/n
n®¥
where an the sequence defined recursively as follows
a1=1 and an+1=n(1-lnan) where ln is the natural logatithm
I can prove (using induction) that the sequence is increasing and not upper bounded but I don't know how to prove that. I'm sorry but I don't know how to write the sequence in the correct form, so I attached a doc file
lim (an)1/n=1 n®¥

2. Setting...

$\displaystyle c_{n}= (a_{n})^{\frac{1}{n}}$

... where the sequence $\displaystyle a_{n}$ is defined as...

$\displaystyle a_{1}=1$, $\displaystyle a_{n+1}=n\cdot (1-\ln a_{n})$

... after easy steps you obtain...

$\displaystyle \ln c_{n}= \frac{\ln (n-1) + \ln (1-\ln a_{n-1})}{n}< \frac{\ln (n-1)}{n}$ (1)

From (1) it derives that...

$\displaystyle \lim_{n \rightarrow \infty} \ln c_{n}=0$

... so that...

$\displaystyle \lim_{n \rightarrow \infty} c_{n}=1$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$

3. thanks $\displaystyle \chi \sigma$ for your answer but i've seen I made a mistake while writing the sequence.
the correct sequence is
$\displaystyle a_{1}=1 \ a_{n+1} = n \cdot(1+\ln a_{n})$
I'm sorry

4. In that case we have to find...

$\displaystyle \lim_{n \rightarrow \infty} c_{n}$ (1)

... where...

$\displaystyle c_{n}= (a_{n})^{\frac{1}{n}}= e^{\frac{\ln a_{n}}{n}}$ (2)

... and the sequence $\displaystyle a_{n}$ is defined as...

$\displaystyle a_{1}=1$, $\displaystyle a_{n+1}= n\cdot (1+\ln a_{n})$ (3)

In this case may help a preliminary test in order to undestand the behavior of the term $\displaystyle \ln a_{n}$ in (2). Value of $\displaystyle \ln a_{n}$ and $\displaystyle \ln n$ computed till to n=1000 are plotted in figure...

Setting...

$\displaystyle \delta_{n}= \ln a_{n} - \ln n$ (4)

... il seems that the sequence $\displaystyle \delta_{n}$ is upper bounded, that means that...

$\displaystyle \lim_{n \rightarrow \infty} \delta_{n}= \delta$ (5)

... where $\displaystyle 2<\delta<3$. In this case is...

$\displaystyle \lim_{n\rightarrow \infty} \frac{\ln a_{n}}{\ln n} = 1$ (6)

... so that...

$\displaystyle \lim_{n \rightarrow \infty} \frac{\ln a_{n}}{n}= \lim_{n \rightarrow \infty} \frac{\ln n}{n}= 0 \rightarrow \lim_{n \rightarrow \infty} c_{n}=1$ (7)

All that however is not yet demonstrated ...

Here there are some value of $\displaystyle \delta_{n}$ with n increasing...

$\displaystyle \delta_{1000}= 2.324429\dots$

$\displaystyle \delta_{10000}= 2.5459131\dots$

$\displaystyle \delta_{50000}=2.6736756\dots$

$\displaystyle \delta_{100000}=2.72369079\dots$

$\displaystyle \delta_{200000}=2.771159376\dots$

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$