Suppose that $\sum\limits_{i=1}^n x_i $ is a convergent series of positive terms that monotonically decrease (that is,$ x_1 \geqslant x_2 \geqslant x_3 \geqslant.. $). Let P denote the set of all numbers that are sums of some (finite or infinite) subseries of $\sum_{n=1}^{\infty}$ xi . Show that P is an interval if and only if $$x_n \leq \sum_{i=n+1}^{\infty}x_i$$ for every integer n,which we call condition(1).
This is the condition of the problem, I guessed part of the solution, but I am as sure of it as I am uncertain.
My solution.
Let P be the interval (a,b), where b>a>0. Since b is the largest sum of all possible subsequences, it includes all elements and equal $\sum\limits_{i=1}^n x_i = S$.
To get the smallest element, you need to take the smallest terms from sequense, since by condition the series converges,that $\lim_{n\to \infty}x_n=0$.This means that to get a we can take terms tending to zero,a=$0$.P=($0$,S).
For any c from the interval we can find a subsequence whose sum is equal to c.Let us assume here that $\exists x_k > \sum_{i=k+1}^{\infty}x_i$.Take $c > x_k$ and $c < x_{k-1}$ (Such as c is exists since the sequence is monotonic).Then for $c-x_k$ exists subsequence which sum equal $x_k$ (it is obvious that all elements of the sequence $< x_k$) ,but $x_k > \sum_{i=k+1}^{\infty}x_i$ .That mean $c-x_k + \sum_{i=k+1}^{\infty}x_i <c$ contradiction.
On the contrary, I don’t really understand how to prove.
Assume that (1) holds, let us show that the range of sums is indeed an interval.
As you described in your part of the proof, if the range of all possible sums is an interval then it is an interval $[0 ,S]$, where $S =\sum_{i = 1}^{\infty}x_i$. Thus we need to show that for every positive number $p < S$ there exists a sequence $\{n_k\}$ of positive integer numbers such that $p = \sum_{k = 1}^{\infty}x_{n_k}$. Let us construct this sequence by the greedy algorithm. On the step $N$ we compare $p$ and $x_N$. If $x_N > p$ we move to the next step otherwise we put $x_N$ into the $\textit{bag}$ and replace $p$ with $p - x_N$.
I claim that the sum of all elements from the $\textit{bag}$ equals $p$. It is clear that this sum does not exceed $p$, let it be denoted by $q$. If $q < p$ thet there is an index $N$ such that $\sum_{k > N}x_k < q - p$. The latter means that every $x_n$ with $n > N$ is in the $\textit{bag}$. Hence we can properly define the index of the last element which is not in the $\textit{bag}$ (the set of such indexes is finite and non-empty), let it be $M$. If $A_{M - 1}$ was the sum of the elements in the $\textit{bag}$ before the step $M$ then $A_{M - 1} + x_M > p$ (we didn't take $x_M$). On the other hand we took every next element, therefore we have $$ q = A_{M - 1} + x_{M + 1} + x_{M + 2} + \ldots < p, $$ which contradicts the assertion (1).
To prove the theorem into the other direction you can notice that if $x_k > p > \sum_{n > k}x_{n}$ then $p$ cannot be represented as a desired sum. ($x_n$ with $n\le k$ do not fit and all others are not enough).