Proving a limit theorem of a sequence (square root)

[shoescreen]

Homework Statement

given the sequence {a_n} converges to A (non zero), show sqrt(a_n) = sqrt(A)

Homework Equations

The Attempt at a Solution

I've tried to expand |sqrt(a_n) - sqrt(A)| as |a_n - A|\|sqrt(a_n) + sqrt(A)| since that gives me the numerator to work with, but I can't figure out how to work with sqrt (a_n).

I'm trying a standard n* epsilon proof, that is finding an n* such that the above quantity is less than any positive real number epsilon.

Thanks for the help!

 

[fmam3]

Suppose $$\lim a_n = A$$ and you want to show $$\lim \sqrt{a_n} = \sqrt{A}$$.

So, right now you need a bound for the denominator. Assuming that we are working with the reals (i.e. no complex numbers for this problem), then see that for the square root to be defined, we must have that $$a_n, A \geq 0$$. Thus, that means we can simplify by writing $$| \sqrt{a_n} + \sqrt{A} | = \sqrt{a_n} + \sqrt{A} \geq \sqrt{A}$$ right?

That should be enough for you to finish the proof...

 

[shoescreen]

hmmm, but the inequality is in the wrong direction. If want to compare |a_n - A|\|sqrt(a_n) + sqrt(A)| with a smaller sequence, I need something bigger in the denominator, and since root(A) <_ root (a_n) + root(A), I can't guarantee anything about the original sequence.

 

[fmam3]

Note that WLOG, we can assume that $$A \ne 0$$ since if $$A = 0$$, then the result is trivial.

In fact, we can even further say that we can assume, WLOG that $$A > 0$$ (note the strict inequality). Then, if we have $$\sqrt{a_n} + \sqrt{A} > A$$, then does that not imply $$\frac{1}{\sqrt{a_n} + \sqrt{A}} < \frac{1}{\sqrt{A}}$$? Can you finish the rest of the proof from here?

 

[shoescreen]

I take back what i said before, the inequality is in the correct direction. It's frustrating, I had all of this before yet trashed it all because of that inequality sign. Ugh, guess i wasn't in math mode.

Thanks for the help!

 

[fmam3]

I take back what i said before, the inequality is in the correct direction. It's frustrating, I had all of this before yet trashed it all because of that inequality sign. Ugh, guess i wasn't in math mode.

Thanks for the help!

No prob :)

 

[skoomafiend]

Note that WLOG, we can assume that $$A \ne 0$$ since if $$A = 0$$, then the result is trivial.

In fact, we can even further say that we can assume, WLOG that $$A > 0$$ (note the strict inequality). Then, if we have $$\sqrt{a_n} + \sqrt{A} > A$$, then does that not imply $$\frac{1}{\sqrt{a_n} + \sqrt{A}} < \frac{1}{\sqrt{A}}$$? Can you finish the rest of the proof from here?

bumping an old post, but i am having trouble with a similar problem.

how would one continue from where he ended off?

thanks!

 

[jbunniii]

Using the equality obtained in the excerpt you quoted, you would write

$$|\sqrt{a_n} - \sqrt{A}| = \frac{|a_n - A|}{|\sqrt{a_n} + \sqrt{A}|} < \frac{|a_n - A|}{\sqrt{A}}$$

Now, can you make the numerator as small as you like by choosing a large enough n? If so, what can you conclude?

 

[skoomafiend]

this is about where I'm getting stuck.
would be able to rephrase that question in any other way?

 

[jbunniii]

You're trying to show that

$$\lim_{n \rightarrow \infty} \sqrt{a_n} = \sqrt{A}$$

What does this statement mean? What is the definition of the limit of a sequence?

 

[skoomafiend]

i am trying to show that for all ε > 0, there is an N such that for all n>=N i'll have,

$$|\sqrt{a_n} - \sqrt{A}| < ε$$

i'm not understanding how to relate this (below) to my epsilon.

$$|\sqrt{a_n} - \sqrt{A}| < \frac{|a_n - A|}{\sqrt{A}}$$

 

[jbunniii]

You know that

$$\lim_{n \rightarrow \infty} a_n = A$$

So given any $\epsilon' > 0$ (I intentionally put a ' on the $\epsilon$ to indicate that it's a different variable) there is an N such that

$$|a_n - A| < \epsilon'$$

for all $n \geq N$.

Now think about what choice of $\epsilon'$ would be useful to get the result you require.

 

[skoomafiend]

$$ε' = ε \sqrt{A}$$

would this work?

thank you for all your help so far.

 

[jbunniii]

Looks promising. Why don't you write out your argument in detail and we'll see if it works.

 

[skoomafiend]

Since,
$$|a_n - A| < ε$$

We have that for,
$$ε' > 0$$
there exists N such that for all n>=N,

$$|\sqrt{a_n} - \sqrt{A}| < ε'$$

so,

$$|\sqrt{a_n} - \sqrt{A}| < ... < \frac{|a_n - A|}{\sqrt{A}} < \frac {ε'}{\sqrt{A}} = ε$$

where
$$ε' = ε\sqrt{A}$$

I feel as though I'm missing some critical parts of the argument.

 

[jbunniii]

You have the right idea, but you didn't arrange the proof quite correctly.

Try starting as follows:

Let $\epsilon > 0$. We seek an $N$ such that whenver $n \geq N$, the following inequality is satisfied:

$$|\sqrt{a_n} - \sqrt{A}| < \epsilon$$

We know that

$$\lim_{n \rightarrow \infty} a_n = A$$

Therefore, given $\epsilon' > 0$, ...

 

[skoomafiend]

i'm definitely having trouble arranging the proof correctly.
the given ϵ′ is for $$| a_n - A| < ϵ'$$ ?

 

[jbunniii]

Right. Given $\epsilon' > 0$, there exists an $N'$ such that

$$|a_n - A| < \epsilon'$$

whenever $n > N'$.

I can choose $\epsilon'$ to be whatever I like, and there is guaranteed to be a corresponding $N'$ that makes the above inequality true. So I choose $\epsilon' = \epsilon\sqrt{A}$...

 

[skoomafiend]

now do i say that for every ϵ > 0, there exists an N such that all n >= N where $$|\sqrt{a_n} - \sqrt{A}| < ϵ = \frac{ϵ'}{\sqrt{2}}$$

 

[jbunniii]

Now use the chain of inequalities from message #8.

 

[skoomafiend]

thank you for all your help. i think i have it now.

 

[skoomafiend]

just one more question, when writing out the formal proof. how/where would i add the case for zero? since that case would need to have epsilon prime set to a different value.

the other thing i was confused about was:
i've been given an e' for a certain N'
but for my proof i have set e' as some multiple of e>0, but for this e>0, there exists another N, s.t all n>=N the inequality holds.
how do i relate the N' and N.

does it have to be specified what N is? or just that it exists given that i KNOW N' exists.

thanks again.

 

[jbunniii]

If you choose $\epsilon' = \epsilon \sqrt{A}$, then you can set $N = N'$. Do you see why?

For the $A = 0$ case, you want to show that if $x_n \rightarrow 0$ then $\sqrt{x_n} \rightarrow 0$.

So, given $\epsilon > 0$, you want to show that there is an $N$ such that

$$|\sqrt{x_n}| < \epsilon$$

whenever $n \geq N$

Now, make a similar (but easier) argument as you did before, using the fact that $x_n \rightarrow 0$.

Hint:

$$|\sqrt{x_n}| < \epsilon$$

if and only if

$$|x_n| < ?$$

 

[skoomafiend]

If you choose $\epsilon' = \epsilon \sqrt{A}$, then you can set $N = N'$. Do you see why?

I don't quite see why that would be true.

 

[jbunniii]

If $n \geq N'$, then

$$|\sqrt{a_n} - \sqrt{A}| = \frac{|a_n - A|}{|\sqrt{a_n} + \sqrt{A}|} < \frac{|a_n - A|}{\sqrt{A}} < \frac{\epsilon'}{\sqrt{A}} = \epsilon$$

which is exactly what you want, right? So it suffices to choose $N = N'$.

 

[jmoen17]

was answered

 

[aatzn]

We know that

$\lim a_n = A$

but this is the same as

$\lim (\sqrt{a_n})^2 = (\sqrt{A})^2$

or

$\lim [(\sqrt{a_n}) (\sqrt{a_n})]= (\sqrt{A}) (\sqrt{A})$

and, for the theorem about the limit of products of sequenze:


$\lim [(\sqrt{a_n}) (\sqrt{a_n})]=(\lim(\sqrt{a_n}))(\lim(\sqrt{a_n}))=(\sqrt{A}) (\sqrt{A})$

than

$\lim \sqrt{a_n} = \sqrt{A}$

QED

NB You can generalize this to other roots!

 

[aatzn]

We know that

$\lim a_n = A$

but this is the same as

$\lim (\sqrt{a_n})^2 = (\sqrt{A})^2$

or

$\lim [(\sqrt{a_n}) (\sqrt{a_n})]= (\sqrt{A}) (\sqrt{A})$

and, for the theorem about the limit of products of sequence:


$\lim [(\sqrt{a_n}) (\sqrt{a_n})]=(\lim(\sqrt{a_n}))(\lim(\sqrt{a_n}))=(\sqrt{A}) (\sqrt{A})$

then

$\lim \sqrt{a_n} = \sqrt{A}$

QED

NB You can generalize to other roots

 

[jbunniii]

and, for the theorem about the limit of products of sequence:$\lim [(\sqrt{a_n}) (\sqrt{a_n})]=(\lim(\sqrt{a_n}))(\lim(\sqrt{a_n}))=(\sqrt{A}) (\sqrt{A})$ then

$\lim \sqrt{a_n} = \sqrt{A}$

What theorem are you using here? It is certainly not the case in general that
$$\lim [(x_n)(x_n)] = (x)(x)\implies \lim x_n = x$$
For example, take $x_n = (-1)^n$ and $x = 1$.

Also, this is a very old thread.

 

[aatzn]

I used the theorem about the limit of products of sequence, as I wrote.
You can use it for sequences with finite limits.
Anyway 1 is not the limit of your sequence, there is no limit for your example.
Do you mean it is a product of two other regular sequences?

 

[jbunniii]

Anyway 1 is not the limit of your sequence, there is no limit for your example.

You do not know that your sequence $\sqrt{a_n}$ has a limit, either. That is what you are trying to prove! Please state precisely what your theorem says.

 

[aatzn]

The theorem says that the limit of a sequence product of two other sequences
(with finite limits!) is the product of their limits.
So, in our case ( $\lim a_n = A$ ), if
$\lim \sqrt{a_n}$
exists finite, than it is
$\sqrt{A}$

 

[jbunniii]

So, in our case ( $\lim a_n = A$ ), if
$\lim \sqrt{a_n}$
exists finite, than it is
$\sqrt{A}$

But you didn't prove that $\lim \sqrt{a_n}$ exists.

 

[aatzn]

You are right.
My weaker demonstration shows how that limit should be, if it exists.
On the other hand, with the previous demonstration not only we know the form of the hypothetical limit, but we are sure that it always exists.
But your sample? what did you mean with post n. #30?
Anyway don’t you think my simple method could be a useful aid for the study of more complicate sequences with nth roots?