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albi · #11 September 10th, 2007, 04:21 PM

Quote:

Originally Posted by kezman

$A_n = n.a_n$
$a_n \to 0$ because is a convergent series.
and $a_{n+1}/a_n<1$
then $\lim \frac{A_{n+1}}{A_n} = \lim \frac{(n+1)}{n}. \frac{a_{n+1}}{{a_n}} < 1$
so $A_n \to 0$

Let us consider the series $\sum x_n = \sum \frac{1}{n^2}$ . We can see that for this one we get: $\lim \frac{x_{n+1}}{x_n} = 1$ .
What is more we know that the series converges.

So for the convergent series $\sum a_n$ you can't conclude that $\lim \frac{a_{n+1}}{a_n}<1$ .

The other part is wrong too....