Definition of Convergence

[burt]

What is the definition of convergence? The reason why I'm asking this question is to understand why [MATH]\frac1x[/MATH] diverges and [MATH]\frac1{x^2}[/MATH] converges.
The [MATH]\lim_{x\to\infty}\frac1x=0[/MATH] as does [MATH]\lim_{x\to\infty}\frac1{x^2}[/MATH]. So why is one convergent and one divergent?

 

[pka]

What is the definition of convergence? The reason why I'm asking this question is to understand why [MATH]\frac1x[/MATH] diverges and [MATH]\frac1{x^2}[/MATH] converges.
The [MATH]\lim_{x\to\infty}\frac1x=0[/MATH] as does [MATH]\lim_{x\to\infty}\frac1{x^2}[/MATH]. So why is one convergent and one divergent?

You need to fire whoever told you that because it is false.
TRUE: \(\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0\) but if \(\displaystyle n\in\mathbb{Z}^+~\&~n>1\) then \(\displaystyle \frac{1}{n^2}<\frac{1}{n}\)
\(\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}} \leqslant \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\) so what do you now think

 

[burt]

You need to fire whoever told you that because it is false.
TRUE: \(\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{1}{n} = 0\) but if \(\displaystyle n\in\mathbb{Z}^+~\&~n>1\) then \(\displaystyle \frac{1}{n^2}<\frac{1}{n}\)
\(\displaystyle \mathop {\lim }\limits_{n \to \infty } \frac{1}{{{n^2}}} \leqslant \mathop {\lim }\limits_{n \to \infty } \frac{1}{n}\) so what do you now think

I don't understand what you are saying is false.

 

[pka]

What is the definition of convergence? The reason why I'm asking this question is to understand why [MATH]\frac1x[/MATH] diverges

THAT is false.

 

[burt]

THAT is false.

I think I understand. 1/x diverges. But, the sum of the series with [MATH]a_n[/MATH] being 1/x diverges.

 

[Dr.Peterson]

What is the definition of convergence? The reason why I'm asking this question is to understand why [MATH]\frac1x[/MATH] diverges and [MATH]\frac1{x^2}[/MATH] converges.
The [MATH]\lim_{x\to\infty}\frac1x=0[/MATH] as does [MATH]\lim_{x\to\infty}\frac1{x^2}[/MATH]. So why is one convergent and one divergent?

I suspect that what you meant to say, or what you were told and didn't realize it, is that the series [MATH]\sum_{k=1}^\infty \frac{1}{k}[/MATH] diverges, or possibly that the improper integral [MATH]\int_1^\infty\frac{1}{x}dx[/MATH] diverges. Is one of those possible?

The sequence [MATH]a_k = \frac{1}{k}[/MATH] converges, as you know.

 

[pka]

I suspect that what you meant to say, or what you were told and didn't realize it, is that the series [MATH]\sum_{k=1}^\infty \frac{1}{k}[/MATH] diverges,.

Well I would never guessed that was what was meant.
Bert, have you studied the harmonic sequence \(\displaystyle H_n~?\) Here it is \(\displaystyle {H_n} = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{{{2^n}}}\)
Here is a standard and usual question:
Prove that for \(\displaystyle n\in\mathbb{Z}^+\) then \(\displaystyle H_n\ge 1+\frac{1}{2^n}\)

 

[Steven G]

What is the definition of convergence? The reason why I'm asking this question is to understand why [MATH]\frac1x[/MATH] diverges and [MATH]\frac1{x^2}[/MATH] converges.
The [MATH]\lim_{x\to\infty}\frac1x=0[/MATH] as does [MATH]\lim_{x\to\infty}\frac1{x^2}[/MATH]. So why is one convergent and one divergent?

You are correct that the terms for both SEQUENCES converge to 0. However, as already pointed out, to get the results you mentioned you are talking about SERIES which is the sum of the elements in the sequence.

There is a test called the convergent test (or is it called the divergent test?). It says that a series will diverge if the limit of the sequence is not 0. More formally it says that a necessary condition for a series to converge is that the limit must be 0.

If you understood the test above clearly then you would know that there can be some series whose limits equal 0 but do diverge.