Quotient Test for Convergence of Series

[unscientific]

Homework Statement

The quotient test can be used to determine whether a series is converging or not. The full description is in the attachment.

Homework Equations

The Attempt at a Solution

( i ) Why must they both follow the same behaviour? Even if p ≠ 0, it says nothing about the type of series Ʃu and Ʃv are! How can one so hastily conclude that if one converges, the other must too and if one diverges the other must too?

( ii ) If p = 0, won't it imply that the limit of u_n = 0 as n approaches infinity? So then series Ʃu must definitely converge. Why did they explain it the other way round (if Ʃv converges, then Ʃu must converge) Isn't Ʃu already known to be converging since limit u_n = 0 ?

( iii ) If p = ∞, won't it imply the limit of v_n = 0 as n approaches infinity? (same case as in part (ii) )

Source: Extracted from Riley, Hobson and Bence " Mathematical Methods for Physics and Engineering " - Chapter 4: Series and Limits

 

[HallsofIvy]

Homework Statement

The quotient test can be used to determine whether a series is converging or not. The full description is in the attachment.

Homework Equations

The Attempt at a Solution

( i ) Why must they both follow the same behaviour? Even if p ≠ 0, it says nothing about the type of series Ʃu and Ʃv are! How can one so hastily conclude that if one converges, the other must too and if one diverges the other must too?
If $u_n/v_n$ goes to $\rho$ we can say that, for n sufficiently large, $u_n< (\rho+1)v_n$ and, if $v_n$ converges, so does $u_n$, by the comparison test. Conversely, if $u_n$ diverges so does $v_n$

( ii ) If p = 0, won't it imply that the limit of u_n = 0 as n approaches infinity? So then series Ʃu must definitely converge. Why did they explain it the other way round (if Ʃv converges, then Ʃu must converge) Isn't Ʃu already known to be converging since limit u_n = 0 ?

No, that does NOT imply that u_n goes to 0. For example, if u_n= n and $v_n= n^2$, $u_n/v_n= 1/n$ which goes to 0 as n goes to infinity but both u_n and v_n go to infinity. The denominator just goes to infinity faster than the numerator.

( iii ) If p = ∞, won't it imply the limit of v_n = 0 as n approaches infinity? (same case as in part (ii) )

No, for the same reason as above. Take $u_n= n^2$, $v_n= n$.

Source: Extracted from Riley, Hobson and Bence " Mathematical Methods for Physics and Engineering " - Chapter 4: Series and Limits

 

[unscientific]

First of all, thank you for replying to my enquiry! It is because of people as dedicated as you are so this site can keep going! (So students like myself can enrich our knowledge!)

However, I'm sorry to trouble you but it would be very kind of you if I could clarify some doubts.

If $u_n/v_n$ goes to $\rho$ we can say that, for n sufficiently large, $u_n< (\rho+1)v_n$ and, if $v_n$ converges, so does $u_n$, by the comparison test. Conversely, if $u_n$ diverges so does [itex]v_n[/tex]

No, that does NOT imply that u_n goes to 0. For example, if u_n= n and $v_n= n^2$, [itex]u_n/v_n= 1/n[/tex] which goes to 0 as n goes to infinity but both u_n and v_n go to infinity. The denominator just goes to infinity faster than the numerator. No, for the same reason as above. Take $u_n= n^2$, $v_n= n$.

( i ) I don't understand how you can take

$u_n< (\rho+1)v_n$

and simply apply the comparison test. I tried to break it down into cases:

Case 1: 0 < p < 1

u_n < v_n (for n tends to infinity) implying:

1a. If v_n converges, so must u_n.
1b. But if u_n converges, it says nothing about v_n since u_n < v_n.

2a. If u_n diverges, so must v_n must diverge.
2b. But if v_n diverges, it says nothing about u_n since u_n < v_n.

Case 2: p > 1
Arrives at the same paradox..(ii)
That was a fantastic counter-example which made me realize that my thinking wasn't thorough enough. If you don't mind please tell me if my understanding is correct:

If p = 0,
It implies that v_n increases at a faster rate than u_n such that eventually p → 0. So since v_n > u_n for all n, then if v_n converges then by comparison test u_n must converge!

(Would this also imply that if u_n diverges, then so must v_n?)

(iii)
If p = ∞,

It implies that u_n increases at a faster rate than v_n such that eventually p → ∞. So since u_n > v_n for all n, then if v_n diverges then by comparison test u_n must diverge!

(Would this imply that if u_n converges, so must v_n?)

 

[Mark44]

Mod note: Moved this thread to the Calculus & Beyond section, which is more appropriate than the Precalc section it was posted in.

 

[tt2348]

Hopefully this might clear up some stuff,

Given $\frac{u_n}{v_n}→ρ$, by definition of convergence... we have for sufficiently large N, $|\frac{u_n}{v_n}-ρ|<ε$
IE, $(ρ-ε)v_n<u_n<(ρ+ε)v_n$
so if we pick ε<ρ for some N, we can guarantee comparison holds :-).
also, the other two problems are equivalent if you consider the reciprocal of the quotient of sequences.

 

[tt2348]

this is ofcourse assuming for non negative sequences, ρ<0 would follow in a similar fashion with some slight adjustments

 

[unscientific]

Hopefully this might clear up some stuff,

Given $\frac{u_n}{v_n}→ρ$, by definition of convergence... we have for sufficiently large N, $|\frac{u_n}{v_n}-ρ|<ε$
IE, $(ρ-ε)v_n<u_n<(ρ+ε)v_n$
so if we pick ε<ρ for some N, we can guarantee comparison holds :-).
also, the other two problems are equivalent if you consider the reciprocal of the quotient of sequences.

I'm sorry but i don't really understand what you just wrote.. first of all what is ε?

Secondly for $|\frac{u_n}{v_n}-ρ|<ε$,

all we arrive at is:

u_n < v_n(p-ε)

or

u_n < v_n(p+ε)

 

[unscientific]

I think I have come up with a way to explain why as long as p ≠ 0 both series must accompany each other with same behaviour (assuming all terms are real and positive).since

$\frac{u_n}{v_n}$ → n as n → ∞, it implies that at

sufficiently large n,

[itex]\frac{u_n}{v_n}[/itex] ≈ [itex]\frac{u_n+1}{v_n+1}[/itex] and is truly reached when n = ∞.

Thus comparing the sequence:
u_n, u_n+1, u_n+2...
v_n, v_n+1, v_n+2...

If the first sequence diverges or converges, the second sequence must therefore follow its pattern by the same ratio of ≈ p!

 

[tt2348]

... Have you not done work with the definition of convergence? And that is definitely not how absolute values work... If I said |x-2|<1... That means 1<x<3...

 

[tt2348]

Look up convergent sequence criteria

 

[unscientific]

... Have you not done work with the definition of convergence? And that is definitely not how absolute values work... If I said |x-2|<1... That means 1<x<3...

damn, my mistake. I put the ' - ' sign to the ratio instead of to ε! My bad!

I tried reading up on convergence: http://en.wikipedia.org/wiki/Limit_(mathematics )

But I don't seem to be making any progress.. not sure why I can't understand it

 

[tt2348]

Here is the definition of convergence for a sequence.
in order for $a_n→a$ it follows that for all ε>0 there exists some N so that for natural numbers n>N, $|a_n-a|<ε$
that is... as we get further out into our sequence, the behavior we'd see is that the sequence will get closer to it's limit, or the distance between the limit and the sequence will get smaller and smaller.
take $|\frac{1}{n}-0|<\epsilon$... since 1/n→0,
that means we can pick any interval and find an N so that it works for all n>N
for instance... if we wanted to consider the case for ε=1/100...
$\frac{1}{n}<\frac{1}{100}⇔n>100=N$
the same way we can do it with the quotient converging to ρ. Halls of Ivy specifically picked for ε=1 in this case.
I suggest though keeping ε<ρ, for positive sequences (makes it a lil prettier)

 

[unscientific]

Here is the definition of convergence for a sequence.
in order for $a_n→a$ it follows that for all ε>0 there exists some N so that for natural numbers n>N, $|a_n-a|<ε$
that is... as we get further out into our sequence, the behavior we'd see is that the sequence will get closer to it's limit, or the distance between the limit and the sequence will get smaller and smaller.
take $|\frac{1}{n}-0|<\epsilon$... since 1/n→0,
that means we can pick any interval and find an N so that it works for all n>N
for instance... if we wanted to consider the case for ε=1/100...
$\frac{1}{n}<\frac{1}{100}⇔n>100=N$
the same way we can do it with the quotient converging to ρ. Halls of Ivy specifically picked for ε=1 in this case.
I suggest though keeping ε<ρ, for positive sequences (makes it a lil prettier)

I get it now, thank you! But I don't see any relation in explaining how the inequality post explains why as long as p≠0 both series will be share the same behaviour..

 

[tt2348]

If (p-e)*v_n<u_n,
(p-e)sum(v_n)<sum(u_n)
So v_n diverge implies u_n diverges. u_n converges implies v_n converges.