Cont: Deeper than primes - Continuation 2

[doronshadmi]

Really? Exactly where does my version use partial sums? For your convenience, here's the definition again. Please just highlight were I make any mention of partial sums whatsoever.

A real-valued sequence (a₁, a₂, a₃, ...) has a limit, L, if and only if for all real numbers ε > 0 there exists a natural number N such that for all n > N, |a_n - L| < ε.​

(a₁, a₂, a₃, ...) is a sequence of partial sums exactly because the index of a_n is some member of N, so |a_n - L| must be > 0.

Moreover here is a quote of yours:

Mathematics gives meaning to infinite series by defining it to be the limit of the sequence of partial sums.

 

[jsfisher]

Are you really so narrowly focused, obsessed actually, on 0.999... that you fail to see the obvious?

One example: (1, 0, 1/2, 0, 1/3, 0, 1/4, 0, 1/5, ...)

Since it diverges it is obviously 0 < |a_n - _L|, so you still have no case.

The sequence obviously converges to 0, and half the elements in the sequence have that limit as their value.

 

[doronshadmi]

The sequence obviously converges to 0, and half the elements in the sequence have that limit as their value.

Realy? since when 0 contributes something to addition?

The sum over (1, 0, 1/2, 0, 1/3, 0, 1/4, 0, 1/5, ...) = the sum over (1, 1/2, 1/3, 1/4, 1/5, ...).

 

[jsfisher]

Realy? since when 0 contributes something to addition?

I said nothing about addition. It is just a sequence. Sequences can have limits. The sequence I presented has a limit of 0.

 

[doronshadmi]

I said nothing about addition. It is just a sequence. Sequences can have limits. The sequence I presented has a limit of 0.

If you speck about the behavior of some function such that it tends from 1 to 0 trough 1/2, 1/3, 1/4, 1/5, ..., then also in this case |a_n - L| must be > 0, exactly as explained in http://www.internationalskeptics.com/forums/showpost.php?p=10649134&postcount=501.

So your definition is limited to trivial cases like (0,0,0,...), (1,1,1,...), (0.34₁₀, 0.34₁₀, 0.34₁₀,...), etc.

 

[jsfisher]

For the record:

If you speck about the behavior of some function such that it tends from 1 to 0 trough 1/2, 1/3, 1/4, 1/5, ..., then also in this case |a_n - L| must be > 0, exactly as explained in

So your definition is limited to the trivial cases like (0,0,0,...), (1,1,1,...), (0.34₁₀, 0.34₁₀, 0.34₁₀,...), etc.

 

[jsfisher]

If you speck about the behavior of some function such that it tends from 1 to 0 trough 1/2, 1/3, 1/4, 1/5, ..., then also in this case |a_n - L| must be > 0, exactly as explained in

So your definition is limited to trivial cases like (0,0,0,...), (1,1,1,...), (0.34₁₀, 0.34₁₀, 0.34₁₀,...), etc.

I have no idea what you are going on about now, but it doesn't follow anything under discussion in any obvious way.

Be that as it may, (1, 0, 1/2, 0, 1/3, 0, 1/4, 0, 1/5, ...) is a sequence; the sequence has a limit; and half the elements of the sequence have the same value as the limit.

Another, even more trivial example would be (1, 1, 1, 1, 1, ...).

 

[doronshadmi]

Be that as it may, (1, 0, 1/2, 0, 1/3, 0, 1/4, 0, 1/5, ...) is a sequence; the sequence has a limit; and half the elements of the sequence have the same [value as the limit.

Even if one of the terms of a given sequence is different from the other terms, for example (1,0,0,0,...), then |a_n - L| > 0.

Another, even more trivial example would be (1, 1, 1, 1, 1, ...).

I have already used this example, which is simply self identity of a given term.

Generally you ignore http://www.internationalskeptics.com/forums/showpost.php?p=10649134&postcount=501 and focused only on the trivial case of self identity, so your definition of limit is very poor.

 

[jsfisher]

Even if one of the terms of a given sequence is different from the other terms, for example (1,0,0,0,...), then |a_n - L| > 0.

Your requirement was for some element of sequence, a_n, to be identical to the sequence's limit. Only one such element was needed. The sequence (1, 0, 0, 0, ...) satisfies your requirement with many such elements.

I have already used this example, which is simply self identity of a given term.

No, it is a sequence. A perfectly normal sequence. And it is a sequence that meets the very criterion you claimed impossible to meet.



You have demonstrated quite admirably your lack of understanding of the meaning of limit. You have demonstrated quite admirably your lack of understanding of sequence. You have demonstrated quite admirably your lack of understanding of series. Will there be more? Or are you satisfied at this point of having completely discredited your qualifications to comment on the value of 0.999...?

 

[doronshadmi]

Your requirement was for some element of sequence, a_n, to be identical to the sequence's limit. Only one such element was needed. The sequence (1, 0, 0, 0, ...) satisfies your requirement with many such elements.

Ok, you have showed that some a_n, which is a partial sum of a given sequence, indeed = L, and in this particular case the definition holds.

But actual definitions must hold in all cases.

So please show it in case of (0.9₁₀, 0.99₁₀, 0.999₁₀, ...), where L = 1.

Also please show it in case of (0.9₁₀, 0.09₁₀, 0.009₁₀, ...), where L = 0.

In both cases no addition is used, but only the notion of partial sums, exactly as written here:

Mathematics gives meaning to infinite series by defining it to be the limit of the sequence of partial sums.

 

[doronshadmi]

Let's look at your definition of limit:

A real-valued sequence (a₁, a₂, a₃, ...) has a limit, L, if and only if for all real numbers ε > 0 there exists a natural number N such that for all n > N, |a_n - L| < ε.

Now look at these parts:

Part 1: ε > 0

Part 2: |a_n - L| < ε

If, for example, the real-valued sequence is (0,0,0,...) then no one of the parts is satisfied, so your definition of limit does not hold for for this case, even if it is obvious that the limit of (0,0,0,...) is 0.

If, for example, the real-valued sequence is (-1,-1,-1,...) then no one of the parts is satisfied, so your definition of limit does not hold for this case, even if it is obvious that the limit of (-1,-1,-1,...) is -1.

etc. ad infinitum.

You also wrote this:

However, that definition comes into play in showing the limit of a particular sequence of partial sums is unity, and that establishes the value for a particular series, and that, in turn, establishes -- rigorously -- that 0.999... is identical to 1.

Let's check your claims by using (0.9₁₀, 0.99₁₀, 0.999₁₀, ...) as the particular sequence of partial sums.

However, that definition comes into play in showing the limit of a particular sequence of partial sums is unity

Really? By using your definition please prove that there is no number that is greater than any term of sequence (0.9₁₀, 0.99₁₀, 0.999₁₀, ...) AND smaller than 1.

Without this proof, also 0.999...₁₀ is greater than any term of sequence (0.9₁₀, 0.99₁₀, 0.999₁₀, ...) even if it is < 1.

 

[jsfisher]

Let's look at your definition of limit:

A real-valued sequence (a₁, a₂, a₃, ...) has a limit, L, if and only if for all real numbers ε > 0 there exists a natural number N such that for all n > N, |a_n - L| < ε.

Now look at these parts:

Part 1: ε > 0

Part 2: |a_n - L| < ε

You have omitted significant parts of the definition. You know, the quantifier that goes with ε, and then that part involving N and its relationship to ε, and let's not forget that part involving n.

 

[jsfisher]

Ok, you have showed that some a_n, which is a partial sum of a given sequence, indeed = L, and in this particular case the definition holds.

Why do you persist in inserting this bogus qualifier? The definition of 'limit' in play here is of a sequence (period), not of a sequence of partial sums derived from a sequence.

If you cannot follow that fundamentally simple idea without overloading it with irrelevant and incorrect baggage, then it is no mystery why your multi-decade battle with Mathematics has yielded you nothing.

 

[doronshadmi]

You have omitted significant parts of the definition. You know, the quantifier that goes with ε, and then that part involving N and its relationship to ε, and let's not forget that part involving n.

This time please do not ignore the following part of post http://www.internationalskeptics.com/forums/showpost.php?p=10650518&postcount=511 by using "the quantifier that goes with ε" for part 1, and "then that part involving N and its relationship to ε," and also "that part involving n" for part 2:

If, for example, the real-valued sequence is (0,0,0,...) then no one of the parts (part1 or part2) is satisfied, so your definition of limit does not hold for this case, even if it is obvious that the limit of (0,0,0,...) is 0.

If, for example, the real-valued sequence is (-1,-1,-1,...) then no one of the parts (part1 or part2) is satisfied, so your definition of limit does not hold for this case, even if it is obvious that the limit of (-1,-1,-1,...) is -1.

etc. ad infinitum.

 

[doronshadmi]

Why do you persist in inserting this bogus qualifier? The definition of 'limit' in play here is of a sequence (period), not of a sequence of partial sums derived from a sequence.

This is not what I have said.

a_n is a particular term of a given sequence, and since it is a particular term (where this term is also some sum), its value is a partial case of that sequence.

 

[doronshadmi]

You also wrote this:

However, that definition comes into play in showing the limit of a particular sequence of partial sums is unity, and that establishes the value for a particular series, and that, in turn, establishes -- rigorously -- that 0.999... is identical to 1

Let's check your claims by using (0.9₁₀, 0.99₁₀, 0.999₁₀, ...) as the particular sequence of partial sums.

However, that definition comes into play in showing the limit of a particular sequence of partial sums is unity

Really? By using your definition please prove that there is no number that is greater than any term of sequence (0.9₁₀, 0.99₁₀, 0.999₁₀, ...) AND smaller than 1.

Without this proof, also 0.999...₁₀ is greater than any term of sequence (0.9₁₀, 0.99₁₀, 0.999₁₀, ...) even if it is < 1.

 

[jsfisher]

This time please do not ignore the following part of post http://www.internationalskeptics.com/forums/showpost.php?p=10650518&postcount=511

I did not ignore any part of that post. I discounted the content because you are clearly confused about the meaning of the definition of 'limit'.

The only thing that needs to be "satisfied" (as you say) is for there to be an N, possibly dependent on ε, that meets requirements of the definition.

Consider the sequence (d₁, d₂, d₃, ...) = (0, 0, 0, ...). I claim that N = 0 satisfies the requirements of the definition to establish that the sequence has a limit of L = 0. Were that not the case, then you should be able to exhibit some ε > 0 and some n > N for which |d_n - L| is not less than ε.

 

[jsfisher]

This is not what I have said.

I quoted you word-for-word. It was exactly what you said.

a_n is a particular term of a given sequence, and since it is a particular term (where this term is also some sum), its value is a partial case of that sequence.

Again, you add baggage that does not below.

a_n is the n-th element of a sequence. Full stop. Period.

 

[doronshadmi]

Consider the sequence (d₁, d₂, d₃, ...) = (0, 0, 0, ...). I claim that N = 0 satisfies the requirements of the definition to establish that the sequence has a limit of L = 0. Were that not the case, then you should be able to exhibit some ε > 0 and some n > N for which |d_n - L| is not less than ε.

"for all real numbers ε > 0" is not satisfied in case of (0,0,0,...) (ε = 0 in this case).

Also "there exists a natural number N such that for all n > N, |a_n - L| < ε" is not satisfied in case of (0,0,0,...) (a_n=0, L=0, ε = 0, and therefore |0 - 0|=0, so |a_n - L| < ε is not satisfied).

 

[doronshadmi]

a_n is the n-th element of a sequence. Full stop. Period.

So what, after all you are the one who wrote:

However, that definition comes into play in showing the limit of a particular sequence of partial sums is unity, and that establishes the value for a particular series, and that, in turn, establishes -- rigorously -- that 0.999... is identical to 1

Actually any n-th element of a sequence is also some sum of a particular sequence of partial sums and this is exactly what "which is a partial sum of a given sequence" means, so again, you have no case (moreover you ignore the rest of post http://www.internationalskeptics.com/forums/showpost.php?p=10650300&postcount=510).

And you also still ignore http://www.internationalskeptics.com/forums/showpost.php?p=10650650&postcount=516.