Problem 36 - Page 2

albi · #11 $Old$ 09-10-2007, 04:21 PM

Quote:

Originally Posted by kezman $View Post$

$A_n = n.a_n$
$a_n \to 0$ 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....

albi · #12 $Old$ 09-10-2007, 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....

ThePerfectHacker · #13 $Old$ 09-10-2007, 07:28 PM

The combinatorics question was answered by Soroban. Here is the solution to the series question. This question comes from my homework on Real Analysis last semester.

For any $\epsilon > 0$ rather consider $\frac{\epsilon}{2} > 0$ . Since the series is convergent by the Cauchy criterion it means $\left| \sum_{k=m}^n a_n \right| = \sum_{k=m}^n a_n < \frac{\epsilon}{2}$ for all $n\geq m\geq N$ . So if $m=N$ then it means $a_N+a_{N+1}+...+a_n < \frac{\epsilon}{2}$ . Since the sequence is non-increasing it means $(n-N)a_n = \underbrace{a_n+...+a_n}_{n-N} \leq a_{N}+...+a_n < \frac{\epsilon}{2}$ for $n\geq 2N$ . Also for $n\geq 2N$ we have $n = 2n - n \leq 2n -2N = 2(n-N)$ . This means $na_n \leq 2a_n(n-N) < \epsilon$ . Q.E.D.

Note: This problem gives us a elegant way to show the divergence of the harmonic series.

albi · #14 $Old$ 09-11-2007, 04:33 AM

Hm, isn't it too early to post the solution, and make the other people have no fun?

I would like to do this more elegant simply taking $\epsilon > 0$ . From Cauchy condition we have (as you noticed): $(k-l)a_l < a_l + a_{l+1} + \ldots + a_k < \epsilon$ . Now we simply take $k = 2n$ and $l = n$ to get $na_n < \epsilon$ .

That means this positive sequence converges to 0.

albi · #15 $Old$ 09-11-2007, 04:37 AM

Quote:

Originally Posted by ThePerfectHacker $View Post$

Note: This problem gives us a elegant way to show the divergence of the harmonic series.

How do you think one should do that?

ThePerfectHacker · #16 $Old$ 09-11-2007, 10:07 AM

Quote:

Originally Posted by albi $View Post$

How do you think one should do that?

Because the harmonic series satisfies that problem but if it were to converge it would imply that $\lim \ n\cdot \frac{1}{n}=0$ which is a contradiction.