Mathematical Analysis Questions

If anyone would be able to check my answers, and/or point me in the right direction on those which are wrong or I'm stuck on, I'd appreciate it.



Claim 1: One such counterexample would be if we let $\displaystyle a_n=1-\frac{1}{n}$ which is a strictly increasing sequence that is bounded above at the value of 1.

Claim 2: $a_n=(-2)^n$ is a counter-example $\forall n \in \mathbb{N}$

Claim 3: Stuck on this as I wasn't sure if $b_n$ must be a constant or not, but I settled on the decision that it cannot be. So if $b_n=\frac{1}{n}$ then $(a_nb_n) \not \rightarrow \infty$ as it would tend to 0 instead, I think.

-----------------------------------------------------------------------------------



I suppose this one leads off the previous part. I think this is not necessarily true but I'm unsure how to prove it, and I haven't made much progress here.

Your counterexample for claim 3 is incomplete. You need to specify an $(a_n)$ sequence to go with that $(b_n)$

If $a_n=n^2$, for example, then your sequence for $b_n$ would still have $(a_nb_n)\to\infty$

For the last bit, rather than look for a counterexample, try and prove it's true.

I suppose this one leads off the previous part. I think this is not necessarily true but I'm unsure how to prove it, and I haven't made much progress here.

So if I say that $a_n=n$ then I assume my counter-example would hold as then $(a_nb_n)={ 1,1,1,1,1,... }$?

I'm not quite sure how to go about the proof here for the general case, but I gave it some sort of go.

Since $b_n>\frac{1}{2}$ we can say that $b_n=\frac{1}{2}+\frac{1}{n}, \forall n \in \mathbb{N}$

Then we know that $(a_n) \rightarrow \infty$ so we can say it is $a_n=n$

Leaving us with $(a_nb_n)=n(\frac{1}{2}+\frac{1}{n})=\frac{1}{2}n+1$ which indeed tends to infinity.

Generalising exponents we can also get $a_n=n^k, b_n=\frac{1}{2}+\frac{1}{n^l}, k \geq 0, l \geq 0 \Rightarrow (a_nb_n)=\frac{n^k}{2}+n^{k-l}$ which is always tending to infinity by the looks of it, but I do not feel like this is a good enough proof as it doesn't seem to branch out to different functions tending to infinity.

Yes.



Well that's a specific example of two sequences that meet the criterion.



Whilst a specific example can sometimes help to give you the idea of how to go about a general proof, I think it will take longer to analyse how/why that's happening here and we could get distracted by details in the specific sequences, than it would be to do it for the general case.

(Note: A sequence isn't necessarily going to have a nice formula defining it.)

We need to start with, what does it mean a sequence tends to infinity (which is what we've been given in the question for a_n)? What's the mathematical definition you've been given?

Qualitatively, we can see that if a sequence goes off to infinity, and we multiply each element of it by at least half, it's still going to go off to infinity, only it's rate of growth may be slower.

PS: Analysis proofs tend to be a shock to the system when first encountered. Can be a bit like you're sailing along nicely, then suddenly lolwtf!!

The definition for a sequence tending to infinity has been given as:

$(a_n) \rightarrow \infty$ if, and only if, $\forall C >0, \exists N\in \mathbb{N}$ such that $\forall n > N, a_n>C$

and this makes sense to me but I am struggling to apply it when introducing, within a proof, a different sequence that has to be strictly above $\frac{1}{2}$.

Since $(a_n) \rightarrow \infty \Rightarrow (a_n)>C, \forall C > 0$ and if $(b_n)>\frac{1}{2}$ then $(a_nb_n) > \frac{1}{2}C$ which still tends to infinity $\forall n > N, N \in \mathbb{N}$

Hope ghostwalker won't mind me taking a stab at clarifying.

...

Which is almost what we want, but not quite. There's a pesky "/2". So, we need to adjust this, somewhere in the point we either go into, or come out of our orginal sequence $(a_n)$

Doesn't D being an arbitrary value not just imply the sequence tends to infinity anyway?

Not in the slightest, and much appreciated. I have added notation/definitions for the sake of clarity of the explanation that wouldn't be part of a standard proof, but I can't see offhand how to avoid this. Please feel free to criticise - think I'm making a bit of a pig's ear of it.


OK, lets take a stab at a proof. Whilst the format below should be sound, it won't actually work, but hopefully you can see how it could be adapted.

With C,D,M,N symbols I defined in my previous post.

Showing $(a_nb_n)$ is unbounded:

We have an arbitrary D > 0.

By the unboundedness of $(a_n)$ and choosing C equal to D

$\exists N, s.t. \forall n>N, a_n > D$

But we're interested in $a_nb_n$, and since $b_n> 0$ we can say

$\exists N, \forall n>N, a_nb_n > Db_n$

and since $b_n > 1/2$

$\exists N, \forall n>N, a_nb_n > D/2$

We now choose our M to equal N.

So, we've shown:

$\forall D>0, \exists M\in\mathbb{N}s.t.\forall n>M, a_nb_n>D/2$

Which is almost what we want, but not quite. There's a pesky "/2". So, we need to adjust this, somewhere in the point we either go into, or come out of our orginal sequence $(a_n)$

Ah that is a good example of how to approach these proofs, thank you! So for the D/2 at the end, would you just multiply both sides by 2 and show that $2(a_nb_n)>D \Rightarrow 2(a_nb_n) \rightarrow \infty \Rightarrow (a_nb_n) \rightarrow \infty$ ?? Other method of getting rid off that "/2", as you say, I think would be to let a different variable equal D/2.

Would this be the correct approach?

Analysis seems like out of this world so far as I am 2 weeks into it aha, trying to get used to all of it is proving to be a hurdle!

[I should emphaszie I'm being pedantic here.

Saying $2(a_nb_2)\to\infty \Rightarrow (a_nb_n)\to\infty$ seems and is obvious, but prove it from the definitions. There are a several simple results that it is useful to use/have, but you really need to prove them first before using them.]

"let a different variable equal D/2." may be heading in the right direction, but I'm not clear what you mean.

I think it best to just say:
We chose D to start with, and then we chose C based on D, but there is no requirement for them to be equal. So, suppose we let C=2D....

,,

So if $\epsilon > 0 \implies \frac{1}{\epsilon} > 0$ and if $0 > C \implies \frac{1}{\epsilon} > C$

I don't know if you mean "if epsilon > 0 then 1/epsilon > 0 and ...", or you mean "if it is true that 'epsilon > 0 implies 1/epsilon > 0' and ..."

As far as the actual proof here, I would focus on why (by intuition) you would expect the result to be true.

That is: if a_n is consistently large and negative, then 1/a_n is consistently small and negative, so 1/|a_n| is small and positive.
You want to show 1/|a_n| < epsilon. Working backwards shows you you will need 0 > 1/a_n > -epsilon (note use of > since a_n and - epsilon are both negative). So you need a_n < -1/epsilon.

if b_n = 1/a_n that would be a counter example for c)