Questions about the formal definition of the limit of a series

CStudent

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Nov 16, 2018
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14
Hey guys, we have started not long ago to learn the term of limes.

So the known definition of the limit of a series goes like that:

If
gif.latex
converges to 0, then
gif.latex
, there exists an
gif.latex
s.t.
gif.latex
gif.latex
gif.latex


I have some elementary questions about the definition, would it be correct if I use the definition a little bit? such that:

* If
gif.latex
gif.latex
s.t.
gif.latex
we get
gif.latex
gif.latex
gif.latex


* If there exists
gif.latex
s.t.
gif.latex
gif.latex
we get
gif.latex
gif.latex
gif.latex


And another type of limes question:

If the series
gif.latex
is defined by
gif.latex
and converges to 0 then
gif.latex
?

Thank you!
 
Hey guys, we have started not long ago to learn the term of limes.

So the known definition of the limit of a series goes like that:

If
gif.latex
converges to 0, then
gif.latex
, there exists an
gif.latex
s.t.
gif.latex
gif.latex
gif.latex


I have some elementary questions about the definition, would it be correct if I use the definition a little bit? such that:

* If
gif.latex
gif.latex
s.t.
gif.latex
we get
gif.latex
gif.latex
L Incorrect
*If there exists
gif.latex
s.t.
gif.latex
gif.latex
we get
gif.latex
gif.latex
gif.latex
Incorrect

And another type of limes question:

If the series
gif.latex
is defined by
gif.latex
and converges to 0 then
gif.latex
? correct
You are really confused!
If \(\displaystyle a_n\) converges to \(\displaystyle L\) then:
1) if \(\displaystyle \varepsilon>0 \)
2) \(\displaystyle \exists N_1\in\mathbb{N}^+\)
3) so that if \(\displaystyle n\ge N_1 \) then \(\displaystyle |a_n-L|<\varepsilon \).
Look at that order of things. Use that order.
 
Last edited:
You are really confused!
If \(\displaystyle a_n\) converges to \(\displaystyle L\) then:
1) if \(\displaystyle \varepsilon>0 \)
2) \(\displaystyle \exists N_1\in\mathbb{N}^+\)
3) so that if \(\displaystyle n\ge N_1 \) then \(\displaystyle |a_n-L|<\varepsilon \).
Look at that order of things. Use that order.
Thank you!
 
Hey guys, we have started not long ago to learn the term of limes.

So the known definition of the limit of a series goes like that:

If
gif.latex
converges to 0, then
gif.latex
, there exists an
gif.latex
s.t.
gif.latex
gif.latex
gif.latex


I have some elementary questions about the definition, would it be correct if I use the definition a little bit? such that:

* If
gif.latex
gif.latex
s.t.
gif.latex
we get
gif.latex
gif.latex
gif.latex


* If there exists
gif.latex
s.t.
gif.latex
gif.latex
we get
gif.latex
gif.latex
gif.latex


And another type of limes question:

If the series
gif.latex
is defined by
gif.latex
and converges to 0 then
gif.latex
?

Thank you!
I am more than a bit confused by this thread. What are you trying to do and why? (By the way, the correct word in English is "limits" rather than "limes," which is the plural of a type of green fruit.) Are you talking about the limit of a series or of a sequence?

You start by giving a definition of convergence of a sequence to zero and seem to want to expand that to a definition of convergence to numbers other than zero, but you also seem to want to formulate equivalent definitions. Or maybe you want to stick with a given definition but deduce consequences that flow from that definition. (Do you mean "use" or "modify": they have different meanings.)

A definition looks like this:

\(\displaystyle \alpha \text { is } \beta \iff \alpha \text { satisfies these conditions.}\)

PKA did not specify what he considered incorrect in your proposed formulations. He is very rigorous so he may have identified several errors in each.

With respect to your first formulation, the fundamental point that seems to me to be missing is that epsilon can be any positive value. You cannot say there is at least one positive value that works.

\(\displaystyle \alpha \implies \exists \ \epsilon > 0\) does not mean \(\displaystyle \epsilon > 0 \implies \beta.\)

\(\displaystyle \text{Given: } 6 > 0 \text { and } a_n = \dfrac{1}{n} \implies\)

\(\displaystyle |a_n - 2| < 6 \text { for all } n \ge 1 \not \implies \displaystyle \lim_{n \rightarrow \infty} a_n = 2.\)
 
I am more than a bit confused by this thread. What are you trying to do and why? (By the way, the correct word in English is "limits" rather than "limes," which is the plural of a type of green fruit.) Are you talking about the limit of a series or of a sequence?

You start by giving a definition of convergence of a sequence to zero and seem to want to expand that to a definition of convergence to numbers other than zero, but you also seem to want to formulate equivalent definitions. Or maybe you want to stick with a given definition but deduce consequences that flow from that definition. (Do you mean "use" or "modify": they have different meanings.)

A definition looks like this:

\(\displaystyle \alpha \text { is } \beta \iff \alpha \text { satisfies these conditions.}\)

PKA did not specify what he considered incorrect in your proposed formulations. He is very rigorous so he may have identified several errors in each.

With respect to your first formulation, the fundamental point that seems to me to be missing is that epsilon can be any positive value. You cannot say there is at least one positive value that works.

\(\displaystyle \alpha \implies \exists \ \epsilon > 0\) does not mean \(\displaystyle \epsilon > 0 \implies \beta.\)

\(\displaystyle \text{Given: } 6 > 0 \text { and } a_n = \dfrac{1}{n} \implies\)

\(\displaystyle |a_n - 2| < 6 \text { for all } n \ge 1 \not \implies \displaystyle \lim_{n \rightarrow \infty} a_n = 2.\)
I will take it under consideration, very comprehensive and beneficial answer.
Thanks.
 
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