1. i equally thanked everybody once! PaulRS' solution is nice and my solution differs from his in the last setp only, where i applied Cesaro's mean, as Moo mentioned, to the sequence

$\displaystyle y_n=\frac{1}{x_{n+1}}-\frac{1}{x_n}.$ that will give you $\displaystyle \lim_{n\to\infty} n x_{n+1}=2$ and thus $\displaystyle \lim_{n\to\infty} (n+1)x_{n+1}=2.$

2. Define $\displaystyle y_n=n x_n$, then $\displaystyle y_{n+1}=(n+1)\ln(1+\frac {y_n}{n})$

If
$\displaystyle \lim_{n\to\infty}y_n$ exists, it implies:
(1)
$\displaystyle \lim_{n\to\infty} \frac {y_n}{n} = 0$
(2) Series $\displaystyle z_1=y_1$, $\displaystyle z_{n+1}=y_{n+1}-y_n$ converges, since $\displaystyle \sum_{i=1}^{n+1} z_i=$$\displaystyle y_{n+1}$.

$\displaystyle z_{n+1}=(n+1) \ln(1+\frac{y_n}{n})-y_n$
$\displaystyle =(n+1) \{\ln(1+\frac{y_n}{n})-\frac{y_n}{n}+\frac{y_n^2}{2n^2}\}+\frac{y_n}{2n} \{2-\frac{n+1}{n}y_n\}$

The first term converges since it is $\displaystyle O(\frac{1}{n^2})$. The second term is $\displaystyle \frac{y_n(2-y_n)}{2}\frac{1}{n}+O(\frac{1}{n^2})$.

For the second term to converge, it requires $\displaystyle \lim_{n\to\infty}y_n(2-y_n)=0$, so $\displaystyle \lim_{n\to\infty}y_n=2$, noting $\displaystyle \lim_{n\to\infty}y_n>0$. This also confirms $\displaystyle \lim_{n\to\infty}y_n$ does exist.

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