# Thread: convergence from comparison test

1. ## convergence from comparison test

Here is the problem:

If $\displaystyle \sum a_{n}$ with $\displaystyle a_{n}>0$ is convergent, then is $\displaystyle \sum \sqrt{a_{n}a_{n+1}}$ always convergent? if not provide a counter example.

Now I don't think that it is also convergent. for example, $\displaystyle 2 \equiv \sqrt{2*2} < \sqrt {2*3}$. If I am right, what would be a counter example in terms of a? or anything else for that matter?

But if it is truly convergent, I know you have to approach this with the fact that a square root is non negative and you have to use a comparison test. but i just don't see how or why
$\displaystyle \sum \sqrt{a_{n}a_{n+1}} < \sum a_{n}$

2. Originally Posted by osudude
Here is the problem:

If $\displaystyle \sum a_{n}$ with $\displaystyle a_{n}>0$ is convergent, then is $\displaystyle \sum \sqrt{a_{n}a_{n+1}}$ always convergent? if not provide a counter example.

Now I don't think that it is also convergent. for example, $\displaystyle 2 \equiv \sqrt{2*2} < \sqrt {2*3}$. If I am right, what would be a counter example in terms of a? or anything else for that matter?

But if it is truly convergent, I know you have to approach this with the fact that a square root is non negative and you have to use a comparison test. but i just don't see how or why
$\displaystyle \sum \sqrt{a_{n}a_{n+1}} < \sum a_{n}$
AMGM!The AM-GM inequality

3. so in application of the amgm, would i just get that $\displaystyle a_{n} < \sqrt{a_{n}a_{n+1}}$ and thus it is divergent from the comparison test??

4. Originally Posted by osudude
so in application of the amgm, would i just get that $\displaystyle a_{n} < \sqrt{a_{n}a_{n+1}}$ and thus it is divergent from the comparison test??
I do not think that you understand the inequality.
$\displaystyle a^2+b^2\ge2ab$.
So in this case $\displaystyle 2\sqrt{a_na_{n+1}}\le a_n+a_{n+1}$

5. Originally Posted by osudude
so in application of the amgm, would i just get that $\displaystyle a_{n} < \sqrt{a_{n}a_{n+1}}$ and thus it is divergent from the comparison test??
Originally Posted by Plato
I do not think that you understand the inequality.
$\displaystyle a^2+b^2\ge2ab$.
So in this case $\displaystyle 2\sqrt{a_na_{n+1}}\le a_n+a_{n+1}$
As Plato pointed out the specific case of the AMGM here is that $\displaystyle \left(\sqrt{x}-\sqrt{y}\right)^2\ge 0$ ( then of course we may expand this to get $\displaystyle \left(\sqrt{x}\right)^2-2\sqrt{xy}+\left(\sqrt{y}\right)^2\implies x+y\ge 2\sqrt{xy}$. I pointed out the general case so that you may note that there is nothing special about there being two terms...except that it is particularly easy to prove the lemma needed.

6. ok so I am still confused. So I understand that

$\displaystyle 2\sqrt{a_na_{n+1}}\le a_n+a_{n+1}$

which equals

$\displaystyle \sqrt{a_na_{n+1}}\le \frac {a_n+a_{n+1}}{2}$

now we know that

$\displaystyle \frac{a_n+a_{n+1}}{2} > \frac{a_{n}+a_{n}}{2} = a_{n}$

. So if

$\displaystyle \frac{a_n+a_{n+1}}{2} > \sqrt{a_na_{n+1}}$

and

$\displaystyle \frac{a_n+a_{n+1}}{2} > \frac{a_{n}+a_{n}}{2} = a_{n}$

Then what does that tell us about the relation between

$\displaystyle \sqrt{a_na_{n+1}}$ and $\displaystyle {a_n}$ ? I mean I want to show, somehow, that $\displaystyle \sqrt{a_na_{n+1}} \leq a_n$ and thus $\displaystyle \sqrt{a_na_{n+1}}$ is convergent by the comparison test

7. Originally Posted by osudude
ok so I am still confused. So I understand that

$\displaystyle 2\sqrt{a_na_{n+1}}\le a_n+a_{n+1}$

which equals

$\displaystyle \sqrt{a_na_{n+1}}\le \frac {a_n+a_{n+1}}{2}$

now we know that

$\displaystyle \frac{a_n+a_{n+1}}{2} > \frac{a_{n}+a_{n}}{2} = a_{n}$

. So if

$\displaystyle \frac{a_n+a_{n+1}}{2} > \sqrt{a_na_{n+1}}$

and

$\displaystyle \frac{a_n+a_{n+1}}{2} > \frac{a_{n}+a_{n}}{2} = a_{n}$

Then what does that tell us about the relation between

$\displaystyle \sqrt{a_na_{n+1}}$ and $\displaystyle {a_n}$ ? I mean I want to show, somehow, that $\displaystyle \sqrt{a_na_{n+1}} \leq a_n$ and thus $\displaystyle \sqrt{a_na_{n+1}}$ is convergent by the comparison test
$\displaystyle \sum\sqrt{a_n a_{n+1}}\le\frac{1}{2}\sum\left\{a_n+a_{n-1}\right\}=a_0+2\sum a_n$