Proving the Truth of \sum^{\infty}_{n=1}|x_{n}|=0

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Homework Statement



Is the proposition

\sum^{\infty}_{n=1}|x_{n}|=0 ⇔\foralln\inIN x_{n}=0

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

Homework Equations


The Attempt at a Solution


I proved the \Leftarrow side of proposition but i could not prove the \Rightarrow side of proposition.
 
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Is it not true that \sum_n |x_n| > x_m for all m?
 
fderingoz said:

Homework Statement



Is the proposition

\sum^{\infty}_{n=1}|x_{n}|=0 ⇔\foralln\inIN 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.
fderingoz said:
If it is true how can we prove that ?

Homework Equations


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.

I am waiting for your answers.
 
fderingoz said:
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.

I am waiting for your answers.

Try a proof by contradiction. Assume one of the a's isn't zero.
 
\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.

I made this proof. What do you think is there any mistake in this proof ?
 
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