# Infinite Sum

## Homework Statement

Is the proposition

$\sum^{\infty}_{n=1}$|x$_{n}$|=0 ⇔$\forall$n$\in$IN x$_{n}$=0

true? If it is true how can we prove that ?

## The Attempt at a Solution

I proved the $\Leftarrow$ side of proposition but i could not prove the $\Rightarrow$ side of proposition.

Is it not true that $\sum_n |x_n| > x_m$ for all m?

Mark44
Mentor

## Homework Statement

Is the proposition

$\sum^{\infty}_{n=1}$|x$_{n}$|=0 ⇔$\forall$n$\in$IN x$_{n}$=0

true?
Is this what you're trying to prove?
$$\sum^{\infty}_{n=1}|x_{n}|= 0 \iff \forall n \in Z,~ x_{n} = 0$$

I wasn't sure what you meant by IN. Also, one symbol you used (⇔) renders as a box in my browser.
Edit: Now it's showing up. That's odd, it didn't before.

Tip: Use one pair of tex or itex tags for the whole expression, rather than a whole bunch of them.
If it is true how can we prove that ?

## The Attempt at a Solution

I proved the $\Leftarrow$ side of proposition but i could not prove the $\Rightarrow$ side of proposition.

Last edited:
Mark44

The proposition which i want to prove is exactly the proposition which you write. I meant natural numbers by IN. Thanks for your suggestions.

Dick
Homework Helper
Mark44

The proposition which i want to prove is exactly the proposition which you write. I meant natural numbers by IN. Thanks for your suggestions.

$\sum\limits_{n=1}^{\infty}|x_{n}|=0\Rightarrow\lim \limits_{n\rightarrow\infty}\sum\limits_{k=1}^{n}|x_{k}|=0$. Since $s_{n}= \sum\limits_{k=1}^{n}|x_{k}|$ is an increasing sequence it can converge only its supremum. So $\sup\limits_{n}s_{n}=0$. Thus $\forall n,\, 0\leq s_{n} = \sum\limits_{k=1}^{n}|x_{k}|\leq 0$ which means $\forall n,\, x_{n}=0$.