Limit (1)

**NonCommAlg** · June 11th 2009, 01:24 PM

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

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

**luobo** · August 7th 2009, 04:19 AM

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

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

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

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

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

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