Converge or diverge plus show why

**Plato** · January 11th 2012, 07:13 AM

Originally Posted by Shizaru

how is it? it's not (1/5)^r. If it was I could see that

$\left|\frac{\sin(r^2)}{5^r}\right|\le\left|\frac{1 }{5^r}\right|=\frac{1}{5^r}$ .

**Shizaru** · January 11th 2012, 08:02 AM

Originally Posted by Plato

$\left|\frac{\sin(r^2)}{5^r}\right|\le\left|\frac{1 }{5^r}\right|=\frac{1}{5^r}$ .

this is what i knew already but dont see why it helps/is neccessary

$\frac{1}{5^{r}} \geq \frac{\sin(r^{2})}{5^{r}}$ without even going into absolutes. i'm asking why absolutes are even being mentioned here. also, why $\frac{1}{5^{r}}$ converges.

^
that's what i was asking about

**Plato** · January 11th 2012, 08:14 AM

Originally Posted by Shizaru

this is what i knew already but dont see why it helps/is neccessary
$\frac{1}{5^{r}} \geq \frac{\sin(r^{2})}{5^{r}}$ without even going into absolutes. i'm asking why absolutes are even being mentioned here. also, why $\frac{1}{5^{r}}$ converges

It is totally basic knowledge.
The geometric series $\sum\limits_{r = 0}^\infty {\frac{1}{{5^r }}}$ converges.

Therefore by the basic comparison test, the series $\sum\limits_{r = 0}^\infty {\left| {\frac{{\sin (r^2 )}}{{5^r }}} \right|}$ converges.

Any series that converges absolutely, converges period.

**Shizaru** · January 11th 2012, 08:20 AM

Once again not what I asked.

I asked why are you going into absolutes in the first place. That is still true without the absolute bit. at what stage is the need to even bring absolutes into it neccesary? since you can use the comparison test without them.

And why does that geometric series converge? I know it is true for anything between -1 and 1 to the power r, but this isnt in that form. i know it does, obviously. but i need to show it.

If I can show 1/5^r converges, i can just use the comparison test with the series in the question, without even going into absolutes.

**Plato** · January 11th 2012, 08:56 AM

Originally Posted by Shizaru

Once again not what I asked.
I asked why are you going into absolutes in the first place. That is still true without the absolute bit. at what stage is the need to even bring absolutes into it neccesary? since you can use the comparison test without them.
And why does that geometric series converge? I know it is true for anything between -1 and 1 to the power r, but this isnt in that form. i know it does, obviously. but i need to show it.
If I can show 1/5^r converges, i can just use the comparison test with the series in the question, without even going into absolutes.

I truly hope someone else will help you.
I taught this material for forty years and must tell you that I have absolutely no idea why you are having such difficulties.

**Shizaru** · January 11th 2012, 03:41 PM

Originally Posted by Plato

I truly hope someone else will help you.
I taught this material for forty years and must tell you that I have absolutely no idea why you are having such difficulties.

Because you haven't once answered the actual question I asked?

The comparison test here works without having to even mention absolutes.
So I asked why you brought this concept into the idea in the first place.

I also asked how you would prove 1/5^r converges, given the material I have clearly states you are NOT allowed to just assume this/take it as a given. the only standard result with geometric series we are allowed to use are in the form of any given value between -1 and 1 to the power r... no reciprocals in this form can be assumed to converge.

The response you gave was a number of things which I already knew, didn't solve what I asked, and aren't actually required for the problem.

You've answered a question I didn't ask. That's the 'problem I'm having', as you say. I appreciate your time and effort to help and I'm not doubting the accuracy of what you posted, but I wish you would see this, is not the actual stage of the problem I am enquiring about.

it's kind of like, a implies b, b implies c. therefore a implies c. however, a already implied c. so why go into b?

**Prove It** · January 11th 2012, 05:31 PM

Originally Posted by Shizaru

Because you haven't once answered the actual question I asked?

The comparison test here works without having to even mention absolutes.
So I asked why you brought this concept into the idea in the first place.

I also asked how you would prove 1/5^r converges, given the material I have clearly states you are NOT allowed to just assume this/take it as a given. the only standard result with geometric series we are allowed to use are in the form of any given value between -1 and 1 to the power r... no reciprocals in this form can be assumed to converge.

The response you gave was a number of things which I already knew, didn't solve what I asked, and aren't actually required for the problem.

You've answered a question I didn't ask. That's the 'problem I'm having', as you say. I appreciate your time and effort to help and I'm not doubting the accuracy of what you posted, but I wish you would see this, is not the actual stage of the problem I am enquiring about.

it's kind of like, a implies b, b implies c. therefore a implies c. however, a already implied c. so why go into b?

It's because if you weren't bringing in the absolute values, then you would need to show that the series is bounded BELOW as well as above, because the series could diverge to $\displaystyle \begin{align*} -\infty \end{align*}$ .

Originally Posted by Shizaru

how is it? it's not (1/5)^r. If it was I could see that

Basic index laws... $\displaystyle \begin{align*} \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \end{align*}$ , so

$\displaystyle \begin{align*} \left(\frac{1}{5}\right)^n = \frac{1^n}{5^n} = \frac{1}{5^n} \end{align*}$

Plato is right, it's a geometric series, which is convergent because the common ratio's absolute value is less than 1.

**Shizaru** · January 12th 2012, 03:10 AM

Originally Posted by Prove It

It's because if you weren't bringing in the absolute values, then you would need to show that the series is bounded BELOW as well as above, because the series could diverge to $\displaystyle \begin{align*} -\infty \end{align*}$ .

Hm.. but, do we not have that $\frac{1}{5^{r}}$ converges, and is greater or equal to $\frac{\sin(r^{2})}{5{^}r}$ for all r, and thus that converges too by the comparison test (regardless of absolutes - i know absolute convergence implies convergence, but that doesnt mean its neccesary to use that property all the time), and that there answers the initial problem?

my qualm is that using absolute convergence is just an unnecesary step here?

Originally Posted by Prove It

Basic index laws... $\displaystyle \begin{align*} \left(\frac{a}{b}\right)^n = \frac{a^n}{b^n} \end{align*}$ , so

$\displaystyle \begin{align*} \left(\frac{1}{5}\right)^n = \frac{1^n}{5^n} = \frac{1}{5^n} \end{align*}$
Plato is right, it's a geometric series, which is convergent because the common ratio's absolute value is less than 1.

thats exactly what i was looking for, thanks a lot. however we are not allowed to assume this as is stated. would i be already just rearranging it like that to show that it is in fact in line with the geometric series rules. also supposing it wasnt a 1 on top, but a 2, or anything else... then this wouldnt be possible would it?

**Prove It** · January 12th 2012, 03:16 AM

Originally Posted by Shizaru

Hm.. but, do we not have that $\frac{1}{5^{r}}$ converges, and is greater or equal to $\frac{\sin(r^{2})}{5{^}r}$ for all r, and thus that converges too by the comparison test (regardless of absolutes - i know absolute convergence implies convergence, but that doesnt mean its neccesary to use that property all the time), and that there answers the initial problem?

my qualm is that using absolute convergence is just an unnecesary step here?

It's not unnecessary, it's VERY necessary, for the reason I stated earlier. By showing absolute convergence, you are showing that the series is bounded BOTH above and below.

It should be clear that $\displaystyle \begin{align*} -|a| \leq a \leq |a| \end{align*}$ , so $\displaystyle \begin{align*} -\sum{|a|} \leq \sum{a} \leq \sum{|a|} \end{align*}$ .

Thats exactly what i was looking for, thanks a lot. however we are not allowed to assume this as is stated. would i be already just rearranging it like that to show that it is in fact in line with the geometric series rules. also supposing it wasnt a 1 on top, but a 2, or anything else... then this wouldnt be possible would it?

If you did not recognise that it was a geometric series, you could use another convergence test, like the ratio test.

If it was some other number on the top...

$\displaystyle \begin{align*} \frac{a}{b^n} = a\left(\frac{1}{b^n}\right) = a\left(\frac{1^n}{b^n}\right) = a\left(\frac{1}{b}\right)^n \end{align*}$

**Shizaru** · January 12th 2012, 04:50 AM

but why do you need to show it is bounded?

the comparison test just says that if the series is smaller than a convergent series for all r, then it converges. and we have that 1/5^r converges, right? so why is it not just as simple as that?

EDIT
actually, looking at wikipedias comparison test article, it says what you are saying, about absolute convergence. however my course notes make no mention of this. you can forgive me this confusion given i am being fed incorrect (or at very least unsufficient...) information from my lecturers.

maybe i should just pay my tuition fees straight to you

**Prove It** · January 12th 2012, 05:26 AM

Originally Posted by Shizaru

but why do you need to show it is bounded?

the comparison test just says that if the series is smaller than a convergent series for all r, then it converges. and we have that 1/5^r converges, right? so why is it not just as simple as that?

In short, the series might diverge, but to $\displaystyle \begin{align*} -\infty \end{align*}$ , not $\displaystyle \begin{align*} +\infty \end{align*}$ . Showing that the series is no greater than some convergent series only shows that it does not diverge to $\displaystyle \begin{align*} +\infty \end{align*}$ . You also need to show that it is no less that some other convergent series to show that it does not diverge to $\displaystyle \begin{align*} -\infty \end{align*}$ . This is why it is handy to prove absolute convergence, because it does both at the same time.

**Shizaru** · January 16th 2012, 11:40 AM

for the third one, ratio test?

i think i have that this diverges but can i ask please, what is the ratio test actually testing? is it saying that if the next term is ever larger than the previous then it diverges? does that mean all convergent series must be decreasing? (if the ratio is greater than 1 then it means the [n+1]th term is larger.)

would it be sufficient to show that for any given value of r (even say, r=1), that the n+1th term is larger than the nth term?

what is the limit of $\frac{\frac{(3(n+1)!)}{(n+1)!^{3}}}{\frac{(3n)!}{( n!)^{3}}}$ ? i've tried manually inputting large n and i am hovering around 26.999.. is the limit 27? i'm certain the limit is 27... can someone please help me with how to show this? the 3s and cubes obviously figure since it's 3^3. i can't make the step with the proof.

would i HAVE to show what the limit is, or just stating it's above 1 work? what if the sequence doesn't have a limit, does the ratio test apply? (i think not...)

**Prove It** · January 17th 2012, 01:45 AM

Originally Posted by Shizaru

for the third one, ratio test?

i think i have that this diverges but can i ask please, what is the ratio test actually testing? is it saying that if the next term is ever larger than the previous then it diverges? does that mean all convergent series must be decreasing? (if the ratio is greater than 1 then it means the [n+1]th term is larger.)

would it be sufficient to show that for any given value of r (even say, r=1), that the n+1th term is larger than the nth term?

what is the limit of $\frac{\frac{(3(n+1)!)}{(n+1)!^{3}}}{\frac{(3n)!}{( n!)^{3}}}$ ? i've tried manually inputting large n and i am hovering around 26.999.. is the limit 27? i'm certain the limit is 27... can someone please help me with how to show this? the 3s and cubes obviously figure since it's 3^3. i can't make the step with the proof.

would i HAVE to show what the limit is, or just stating it's above 1 work? what if the sequence doesn't have a limit, does the ratio test apply? (i think not...)

Why not try simplifying before taking the limit?

$\displaystyle \begin{align*} \frac{\frac{[3(r+1)]!}{[(r + 1)!]^3}}{\frac{(3r)!}{(r!)^3}} &= \frac{(3r + 3)!(r!)^3}{(3r)![(r + 1)!]^3} \\ &= \frac{(3r+3)(3r+2)(3r+1)(3r)!(r!)^3}{(3r)![(r + 1)r!]^3} \\ &= \frac{(3r+3)(3r+2)(3r+1)(r!)^3}{(r+1)^3(r!)^3} \\ &= \frac{(3r+3)(3r+2)(3r+1)}{(r + 1)^3} \\ &= \frac{27r^3 + 54r^2 + 33r + 6}{r^3 + 3r^2 + 3r + 1} \\ &= \frac{27 + \frac{54}{r} + \frac{33}{r^2} + \frac{6}{r^3}}{1 + \frac{3}{r} + \frac{3}{r^2} + \frac{1}{r^3}}\end{align*}$

I also get 27 when you make this go to infinity.

This is enough to show that the series diverges.

**Shizaru** · January 17th 2012, 02:19 AM

Originally Posted by Prove It

Why not try simplifying before taking the limit?

$\displaystyle \begin{align*} \frac{\frac{[3(r+1)]!}{[(r + 1)!]^3}}{\frac{(3r)!}{(r!)^3}} &= \frac{(3r + 3)!(r!)^3}{(3r)![(r + 1)!]^3} \\ &= \frac{(3r+3)(3r+2)(3r+1)(3r)!(r!)^3}{(3r)![(r + 1)r!]^3} \\ &= \frac{(3r+3)(3r+2)(3r+1)(r!)^3}{(r+1)^3(r!)^3} \\ &= \frac{(3r+3)(3r+2)(3r+1)}{(r + 1)^3} \\ &= \frac{27r^3 + 54r^2 + 33r + 6}{r^3 + 3r^2 + 3r + 1} \\ &= \frac{27 + \frac{54}{r} + \frac{33}{r^2} + \frac{6}{r^3}}{1 + \frac{3}{r} + \frac{3}{r^2} + \frac{1}{r^3}}\end{align*}$

I also get 27 when you make this go to infinity.

This is enough to show that the series diverges.

why not you ask? well obviously because i'm not as smart as you to think of all this ^^^^ thank you very much!

Well, it's the first of this type of question i've attempted so I don't know the strategy. I tried to take the limit but just got infinity over infinity which wasn't right. I know that given this limit of 27, the ratio test shows it to diverge. It's just showing what I knew, that the limit of the sequence is 27.

cancelling the factorial factors like that is genius stuff... where do you get the vision to be able to see this things (I have no problem with solving, or intution, I just can never find this strategy)?

EDIT

i'm trying to work through that limit proof. on the second line on the right at the end you put [(r+1)!r!]^3 but where did that extra r!^3 come from?

**Prove It** · January 17th 2012, 02:25 AM

Originally Posted by Shizaru

why not you ask? well obviously because i'm not as smart as you to think of all this ^^^^ thank you very much!

Well, it's the first of this type of question i've attempted so I don't know the strategy. I tried to take the limit but just got infinity over infinity which wasn't right. I know that given this limit of 27, the ratio test shows it to diverge. It's just showing what I knew, that the limit of the sequence is 27.

cancelling the factorial factors like that is genius stuff... where do you get the vision to be able to see this things (I have no problem with solving, or intution, I just can never find this strategy)?

One word... Experience

Keep practicing and it will become second nature

Thread: Converge or diverge plus show why

LinkBack

Thread Tools

Display

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Re: Converge or diverge plus show why

Similar Math Help Forum Discussions

how to you show the series sigma n^(3/n) converge or diverge?

Give examples to show that the following series may converge or diverge

Converge or diverge

Converge of Diverge...why?

converge or diverge

Search Tags