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albi · #12 September 10th, 2007, 04:50 PM

1. Suppose there exists k where $a_k$ . But then $a_n \leq a_k$ for $n > k$ . So $\lim a_n \leq a_k < 0$ . Thus the corresponding series does not converge which contradicts assumptions.

2. So now we know that $a_n$ is a positive sequence. So let us assume that $a_n \geq \frac{\epsilon}{n}$ . By the 'comparison test' (or whatever it is called in English) we have that series $\sum a_n$ does not converge which again contradicts the assumptions.

So we have $a_n < \frac{\epsilon}{n}$ for some n, end arbitrary chosen $\epsilon > 0$ . What is more there are infinitly many elements with this property.

3. Now let us assume that $n a_n$ converges. From 2. we have subsequence $n_k$ where $a_{n_k} < \frac{\epsilon}{n_k}$ .

So we get $0 \leq \lim n a_n = \lim n_k a_{n_k} \leq \epsilon$

Because $\epsilon$ is arbitrary we can put $\epsilon \rightarrow 0$ . Then we get:

$\lim n a_n = 0$

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So now we should prove that $n a_n$ converges....

Or maybe go to sleep....