A question about series

Aldarion · #1 $Old$ June 3rd, 2009, 08:20 AM

I've got this question I can't even think how to begin answering...

Let $a_1, a_2, ...$ be a monotonically non-increasing sequence of positive numbers, so that the series $\sum_{n=1}^{\infty} a_n$ converges. Let S be the set of all numbers obtainable as $\sum_{j=1}^{\infty} a_{n_j}$ for some series $n_1 < n_2 < n_3 < ...$ .
Show that S is an interval iff for all $n \in N$ ,
$a_n \le \sum_{j=n+1}^{\infty} a_j$ .

For which $r > 0$ does $\sum r^n$ satisfy the condition?

Thanks in advance...

Media_Man · #2 $Old$ June 4th, 2009, 01:23 PM

Quote:

Originally Posted by Aldarion $View Post$

$a_n \le \sum_{j=n+1}^{\infty} a_j$ .

For which $r > 0$ does $\sum r^n$ satisfy the condition?

The last part of this question is not so bad...

For what value of r does the following hold: $r^n \leq \sum_{j=n+1}^{\infty} r^j$ ?

Solution: Let $r^n \leq \sum_{j=n+1}^{\infty} r^j =\frac{r^{n+1}}{1-r}$ . Solving, $r \geq \frac12$ . And as you know, $\sum_{n=0}^{\infty} r^n$ converges iff $r<1$ . So the answer to this part of the question is $r\in[\frac12,1)$

For the first part of the question, consider this linear mapping... $\phi: S\rightarrow[0,1]$ , $\phi(s)=\phi(\sum_{i\in A}a_i)=\sum_{i\in A}\frac1{2^i}$ for some $A\subset\mathbb{N}$

For example, $\phi(a_1+a_3+a_5+a_7+...)=.10101010101_2=\frac23$

Aldarion · #3 $Old$ June 6th, 2009, 10:55 PM

I'm sorry, but I still don't understand how does this linear mapping solve the first part of the question.

Thanks for the second part though!

Laurent · #4 $Old$ June 7th, 2009, 06:36 AM

Quote:

Originally Posted by Aldarion $View Post$

Let $a_1, a_2, ...$ be a monotonically non-increasing sequence of positive numbers, so that the series $\sum_{n=1}^{\infty} a_n$ converges. Let S be the set of all numbers obtainable as $\sum_{j=1}^{\infty} a_{n_j}$ for some series $n_1 < n_2 < n_3 < ...$ .
Show that S is an interval iff for all $n \in N$ ,
$a_n \le \sum_{j=n+1}^{\infty} a_j$ .

This is an "if and only if", so there are two parts.

The "only if" part: suppose there exists $n_0\in\mathbb{N}^*$ such that $a_{n_0} > \sum_{j=n_0+1}^{\infty} a_j$ . Then the following numbers are in $S$ : $\alpha=\sum_{n=1}^{n_0-1} a_n + a_{n_0}$ and $\beta=\sum_{n=1}^{n_0-1} a_n + \sum_{n=n_0+1}^\infty a_n$ , and $\alpha>\beta$ . But the real numbers in $(\beta,\alpha)$ do not belong to $S$ (I let you prove this fact), so that $S$ is not an interval.

N.B. : I'm not sure whether $\alpha$ is in $S$ , it depends what is accepted as a subsequence. But it doesn't matter for this proof since $s=\sum_{n=1}^\infty a_n$ is surely in $S$ , and it is greater than $\alpha$ . We have $0\in S$ and $s\in S$ , and $(\beta,\alpha)\subset[0,s]$ hence, to prove that $S$ is not an interval, it suffices to show that some element of $(\beta,\alpha)$ is not in $S$ .

The "if" part: suppose for all $n$ , $a_n \le \sum_{j=n+1}^{\infty} a_j$ . Obviously, since $a_n\ge 0$ , the endpoints of $S$ are 0 and $s=\sum_{n=1}^\infty a_n$ . Let $0<x<s$ . We need to define a subsequence $(a_{n_i})_i$ such that $\sum_{i=1}^\infty a_{n_i}=x$ . Procede as follows: put $a_1,\ldots,a_k$ in the subsequence, where $k$ is the greatest index such that $\sum_{i=1}^k a_i< x$ . Then wait for the first $l>k$ such that $\sum_{i=1}^k a_i + a_l<x$ , and put $a_l,a_{l+1},\ldots,a_m$ in the subsequence, where $m$ is the greatest index such that $\sum_{i=1}^k a_i + \sum_{i=l}^m a_i <x$ . And so on... We try to approach $x$ from below using terms in the series. You have to show that this procedure makes sense, and that the subsequence you are making sums to $x$ .

That's the plan.