1. ## A question about series

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?

2. Originally Posted by Aldarion
$a_n \le \sum_{j=n+1}^{\infty} a_j$ .

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

3. 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!

4. Originally Posted by Aldarion
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. 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, 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 . 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.