Identifying a convergent/divergent series (1 Viewer)

[Aysce]

Was watching a few videos and came across this:

If <a href="http://www.codecogs.com/eqnedit.php?latex=\lim_{n \to \infty} a_{n} \neq 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\lim_{n \to \infty} a_{n} \neq 0" title="\lim_{n \to \infty} a_{n} \neq 0" /></a> the series is divergent.

I understand this part because as you go through more of the later terms of the series and their sums are not approaching 0, then the terms aren't too small and will add up to infinity.

What I don't understand is this:

If <a href="http://www.codecogs.com/eqnedit.php?latex=\lim_{n \to \infty} a_{n} = 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\lim_{n \to \infty} a_{n} = 0" title="\lim_{n \to \infty} a_{n} = 0" /></a> the series is convergent OR divergent.

I can see why it can be considered convergent, but why can it possibly be divergent?

Thank you!

 

[deswa1]

Look at the harmonic series (particularly the comparison test under the divergence heading): http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

 

[Aysce]

Look at the harmonic series (particularly the comparison test under the divergence heading): http://en.wikipedia.org/wiki/Harmonic_series_(mathematics)

Sorry deswa, I still don't understand

I mean if the sum of nth terms of a series approaches 0 as n -> infinity, how can it possibly diverge? It's not heading to either infinity or negative infinity?

 

[Sy123]

Sorry deswa, I still don't understand

I mean if the sum of nth terms of a series approaches 0 as n -> infinity, how can it possibly diverge? It's not heading to either infinity or negative infinity?

This was pretty mind blowing for me as well.

Consider the series:



This series infact diverges (although very slowly), to see why this could be.

First sketch a graph of y=1/x

Draw the points,

(1,1/1)
(2,1/2)
(3,1/3)
(4,1/4)
(5,1/5)

and so on.

Draw rectangles all with width 1. However make the rectangles underneath the curve.

So we can already establish that:



Lets evaluate the integral on the LHS



So we can see that the harmonic series is divergent despite the summation term converging to zero.

 

[D94]

Sorry deswa, I still don't understand

I mean if the sum of nth terms of a series approaches 0 as n -> infinity, how can it possibly diverge? It's not heading to either infinity or negative infinity?

That's not what you wrote though. You wrote the a_n term as n approaches infinity - that's different from the series.

For example: sum of 1/k from k = 1 to n, illustrates your question. The 1/k^th term as n approaches infinity is zero. But by the p-series test, it diverges.

We can tell by considering the sum of partitions of 1 unit wide and f(x) high, being greater or equal to the integral of 1/x from 1 to n+1. This gives us ln(n+1) and since the logarithmic curve is the slowest growing curve as it approaches infinity, ln(x+1) -> infinity as n approaches infinity. Therefore, the series approaches infinity as n approaches infinity.

 

[asianese]

Good Aysce.

 

[Trebla]

Was watching a few videos and came across this:

If <a href="http://www.codecogs.com/eqnedit.php?latex=\lim_{n \to \infty} a_{n} \neq 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\lim_{n \to \infty} a_{n} \neq 0" title="\lim_{n \to \infty} a_{n} \neq 0" /></a> the series is divergent.

I understand this part because as you go through more of the later terms of the series and their sums are not approaching 0, then the terms aren't too small and will add up to infinity.

What I don't understand is this:

If <a href="http://www.codecogs.com/eqnedit.php?latex=\lim_{n \to \infty} a_{n} = 0" target="_blank"><img src="http://latex.codecogs.com/gif.latex?\lim_{n \to \infty} a_{n} = 0" title="\lim_{n \to \infty} a_{n} = 0" /></a> the series is convergent OR divergent.

I can see why it can be considered convergent, but why can it possibly be divergent?

Thank you!

It's not at simple as that. Without going into technicalities (like monotone and dominated convergence) the individual terms in the sequence need to converge 'fast enough' for the series to converge.

 

[Peter-PF7]

This is HSC stuff...!? :S

 

[Aysce]

It's not at simple as that. Without going into technicalities (like monotone and dominated convergence) the individual terms in the sequence need to converge 'fast enough' for the series to converge.

Oh alright, I'll just have to search more I guess. It was an introductory patrickjmt video so I guess it's not meant to really go into that much detail :/

 

[Sy123]

Oh alright, I'll just have to search more I guess. It was an introductory patrickjmt video so I guess it's not meant to really go into that much detail :/

For some topic patrickJMT doesn't really go into much detail in general, I recommend khanacademy

 

[Shadowdude]

This is HSC stuff...!? :S

Nah, first year uni stuff. Not sure why it's here.

 

[Trebla]

Nah, first year uni stuff. Not sure why it's here.

Agreed

 

[seanieg89]

Incidentally, this is a great example of why we need rigorous definitions and proof in mathematical analysis. The high school notion of a limit is vague and imprecise, which makes false statements like this plausible.