Jam'
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Hi

I've been looking at the definition of a null sequence and don't quite understand it.

My understanding of it leads to the notion that every sequence is a null sequence which is obviously false.

This is my reasoning... Please help me and help me correct my mistakes. Name:  DSC_0039.jpg
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I don't quite understand the mathematical statement with the three quantifiers is the issue I think
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ghostwalker
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(Original post by Jam')
...
A null sequence is one that converges to zero.

In terms of the definition. Choose any distance from zero (that's our epsilon), then there is a point in the sequence, N, beyond which (n>N) all values, f(n), in the sequence lie within epsilon of zero. |f(n)| < epsilon.
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DFranklin
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(Original post by Jam')
I don't quite understand the mathematical statement with the three quantifiers is the issue I think
Yes, this is the problem.

The definition says:

... \forall n \in \mathbb{N}, (n \ge N \implies |f(n)| &lt; \epsilon ).

Rewriting this to be a bit wordier, we get: "for all natural numbers n, the statement "n \ge N" implies |f(n)| < epsilon.

Your interpretation is more like: "the statement "for all natural numbers n, n \ge N" implies |f(n)| < epsilon". (If you wrote this with quantifies it wouldn't look like the definition, it would be more like: ... (\forall n \in \mathbb{N}, \,n \ge N) \implies |f(n)| &lt; \epsilon )).

As far as understanding the definitions goes, it's probably best not to think too much in terms of the quantifier symbols - they save writing, but they don't help clarity.

Instead:

f(n) is null if given \epsilon &gt; 0 we can find N such that whenever n is an integer > N, we have |f(n)| &lt; \epsilon.
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Jam'
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(Original post by DFranklin)
Yes, this is the problem.

The definition says:

... \forall n \in \mathbb{N}, (n \ge N \implies |f(n)| &lt; \epsilon ).

Rewriting this to be a bit wordier, we get: "for all natural numbers n, the statement "n \ge N" implies |f(n)| < epsilon.

Your interpretation is more like: "the statement "for all natural numbers n, n \ge N" implies |f(n)| < epsilon". (If you wrote this with quantifies it wouldn't look like the definition, it would be more like: ... (\forall n \in \mathbb{N}, \,n \ge N) \implies |f(n)| &lt; \epsilon )).

As far as understanding the definitions goes, it's probably best not to think too much in terms of the quantifier symbols - they save writing, but they don't help clarity.

Instead:

f(n) is null if given \epsilon &gt; 0 we can find N such that whenever n is an integer > N, we have |f(n)| &lt; \epsilon.
Ohhhh thanks I think I sort of get this. Is it saying therefore that as epsilon tends to 0, f(n) n-> infinity?

And because |f(n)| < Epsilon => lim n-> infinity f(n) = 0? Hence why the sequence converges?
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Jam'
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(Original post by ghostwalker)
A null sequence is one that converges to zero.

In terms of the definition. Choose any distance from zero (that's our epsilon), then there is a point in the sequence, N, beyond which (n>N) all values, f(n), in the sequence lie within epsilon of zero. |f(n)| < epsilon.
Thanks the picture is much clearer thanks to you and Dfranklin
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