Prove that set contains maximum

Question

Let $$\sum_{k=1}^{\infty}x_k$$ be positive and converging series. Prove that the set $A$ containing the terms of the series has maximum.

According to a theorem we proved in class, a positive series converges if and only if $\exists M>0$ such that $S_n\leq M \forall n\in \mathbb{N}$.

Thus, we get $$M\geq S_n=x_1+x_2+...+x_n\geq x_n \forall n\in\mathbb{N}.$$ However, now I also need to show that $$M\in A=\{x_k :x_k\in(x_{k})\}.$$Unfortunately, I am out of ideas. Should I show that $M\not \in A$ would contradict my assumptions?

Answer 1

Since the series converges, we have $x_k \to 0$.

Let $j$ the first index such that $x_j>0$. If $j$ does not exists, then $x_k=0~~ \forall k$ and $\sup A = 0 \in A$. Since all terms are positives and $x_k \to 0$, there exists $N$ such that :

$$\forall n \geq N, x_k\leq x_j.$$

Then $\sup A = \max\lbrace x_1, \dots,x_N\rbrace$ which is obviously in $A$.

Answer 2

Other way :)

The infinite sum is positive then it's necessary the existance of some $x_k$ positive (otherwise the series would be negative). Take $x_s$ the first positive term.

From the convergence of the series we have $\lim x_k=0$. Then there exists $n_0 \in N$ such that for $n>n_0$ and $\varepsilon<x_s$ we have $$x_n \in (-\varepsilon, \varepsilon) $$ for all $n>n_0$.

So $\{ x_1, \cdots, x_{n_0}\}=K$ is a finite set which contains the positive term $x_s$.

K has an maximum $x_M \geq x_s$. (because it is a finite set, and all finite set have a maximum)

$x_M$ is the maximum of $(x_k)$ because the remaining terms of $x_k$ are in the interval $(-\varepsilon, \varepsilon) $ and are lesser than $x_S \leq X_M$, so $X_M$ is really the maximum of the sequence $(x_k)$.