question in limits

[lqg]

1) let a1,b1>0 b_n+1=sqrt(a_n*b_n) a_n+1=(a_n+b_n)/2

prove the existence of lim an as n->inf and lim bn as n->inf, and prove that they are equal?

 

i figure i need to use epsilon ofcourse but the problem is with N, what its quantity should be with regard to epsilon and the variables here.

 

thanks in advance.

[matt grime]

No way is this an epsilon type question.

 

If a_n a nd b_n converge then showing the limits are as required can be done by taking limits in the defining equations.

 

Secondly, just remember that any increasing sequence bounded above or decreasing sequence bounded below converges.

[lqg]

okay, forgot to look on monotonic series, thanks for the tip.

[lqg]

here iv'e got another four questions, any pointers or hints would be appreciated.

in the attached doc file.

I have four questions about limits.doc

[matt grime]

Oh look, a word document. You should avoid using word to do maths since it is rubbish at it (and most other things). I work in Linux, Unix, and OS X, I might be able to decode that propietary encoded document on one of them. Try pdf, or ps, or just type them out here; it isn't very hard to learn latex to the required standard for this forum.

 

Here, let me do it for you using cut and paste.

 

Decoded in oowriter:

 

I have four questions about limits, or rather trying to see if my approach/solution is correct.

1) the are given two series: {a_n}, {b_n} in both of them n=1 till infinity, when {b_n} increases b_n+1>b_n up to infinity, and lim (an+1-an)/(bn+1-bn)=G as n approaches infinity.

Prove that lim (an/bn) as n approaches infinity exist and that it equals G.

 

2) define: a1=1 a2=2, an+2=sqrt(anan+1), calculate lim an as n approaches inf.

 

3)let a1=0 and for n>=1; a_2n+1=0.5+a_2n; a_2n=(a_2n-1)/2, calculate lim sup a_n, and lim inf a_n.

 

4)calculate lim (1-1/(n^2-4))^(3n^2+5) as n approaches infinity.

 

My apparoches/solutions:

 

1)every lim will apparoach infinity:

lim (a_n+1-a_n)=G(lim(b_n+1-b_n))

lim a_n=lim a_n+1-G(lim(b_n+1-b_n))=a-G(lim(b_n+1-b_n)=g

lim b_n=G(lim(b_n+1)=f

if an->g and b_n->f and b_n doesn’t equal 0 then for all n natural (a_n/b_n)->(g/f) therefore there exist a limit.

 

Lim (a_n/b_n)= Lim |a_n/b_n|=|(a-Glim(b_n+1-b_n))/Glim(b_n+1)|=

=a/(Glim(b_n+1))-(lim(b_n+1-b_n)/lim(b_n+1)= a/(Glim(b_n+1))-1+lim(b_n/b_n+1)=lim(a/(G(b_n+1))-1 as you can see, not equal G, so where did I blow it?

 

2)here I know it equal the famous sqrt(2*sqrt(2*sqrt2….)

but how do I prove it.

 

3)limsup a_n=1; lim inf a_n=0, am I right?

4)I got e^3.

 

 

you didn't even appear to use any maths typesetting, so why the cumbersome file format?

 

 

i'll put my indented comments in another copy:

 

1) the are given two series: {a_n}, {b_n} in both of them n=1 till infinity

 

a series is a sum of terms, these are presumably sequences. that the index goes over n in N is automatic

 

when {b_n} increases b_n+1>b_n up to infinity, and lim (an+1-an)/(bn+1-bn)=G as n approaches infinity.

Prove that lim (an/bn) as n approaches infinity exist and that it equals G.

let me latex up the next one click to see the code

 

2) define:

[math]a_1=1 a_2=2, a_{n+2}=\sqrt{a_na_{n+1}}[/math]

 

at least I think that is how it ought to read. let me retype the next one as well so it is easier to read, let me know if i get it right.

 

, calculate lim an as n approaches inf.

 

3)let a_1=0 and for n>=1; a_{2n+1}=0.5+a_{2n}; a_{2n}=(a_{2n-1})/2, calculate lim sup a_n, and lim inf a_n.

 

4)calculate lim (1-1/(n^2-4))^(3n^2+5) as n approaches infinity.

 

My apparoches/solutions:

 

1)every lim will apparoach infinity:

lim (a_{n+1}-a_n)=G(lim(b_{n+1}-b_{n}))

 

no, you can't do that, those limits might not exist (ie they may be infinite ans so you can't treat them as numbers.

 

lim a_n=lim a_n+1-G(lim(b_n+1-b_n))=a-G(lim(b_n+1-b_n)=g

lim b_n=G(lim(b_n+1)=f

if an->g and b_n->f and b_n doesn’t equal 0 then for all n natural (a_n/b_n)->(g/f) therefore there exist a limit.

 

Lim (a_n/b_n)= Lim |a_n/b_n|=|(a-Glim(b_n+1-b_n))/Glim(b_n+1)|=

=a/(Glim(b_n+1))-(lim(b_n+1-b_n)/lim(b_n+1)= a/(Glim(b_n+1))-1+lim(b_n/b_n+1)=lim(a/(G(b_n+1))-1 as you can see, not equal G, so where did I blow it?

 

2)here I know it equal the famous sqrt(2*sqrt(2*sqrt2….)

but how do I prove it.

 

3)limsup a_n=1; lim inf a_n=0, am I right?

4)I got e^3.

 

the last one sounds plausible, though it ought to be -3 I think.

 

for 2 show it is bounded and increasing, or decreasing and hence result.

[the tree]

Oh look, a word document. You should avoid using word to do maths since it is rubbish at it (and most other things)

As far as I'm aware the only decent WYSIWYG for maths is OO-Math which only seems to be supported by OO. So I figure the best thing would just have been to post it and use the Latex feature.

[lqg]

what about the first question, any tips there?

 

so far, youv'e encrypted it very well.

 

i don't have time to learn latex, if i had you bet your money i would post by latex.

[the tree]

Urm, what the hell, you appear to have written: [math]\lim_{n \to \infty}\frac{a_{n}+1-a_{n}}{b_{n}+1-b_{n}}=G[/math] where a_n is canceled out by -a_n and the same happens on the denominator so you wind up with [math]\lim_{n \to \infty}\frac{1}{1}=G[/math]

 

And learning LaTeX doesn't take much time at all if you only learn what you need as you need it.

[lqg]

what i have typed is:

lim a_n+1-a_n/b_n+1-b_n=G

[Dave]

As far as I'm aware the only decent WYSIWYG for maths is OO-Math which only seems to be supported by OO. So I figure the best thing would just have been to post it and use the Latex feature.

 

Oh, I don't know. LyX is pretty good, and does proper LaTeX rendering.

[the tree]

what i have typed is:

lim a_n+1-a_n/b_n+1-b_n=G

So [math]\lim_{n \to \infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}[/math]? That'd make a lot more sense, sorry for not getting it first time round.

[matt grime]

As far as I'm aware the only decent WYSIWYG for maths is OO-Math which only seems to be supported by OO. So I figure the best thing would just have been to post it and use the Latex feature.

 

 

Well, modulo the fact that there is no such thing as a decent WYSIWYG editor for latex at all, what about winedit for the windows based, lyx, and the many others out there.

 

If you want to make latex documents use emacs or (g)vi(m). or just a plain text editor.

 

however, i don't see what this WYSIWYG debate or otherwise has to do with anything. I wasn't even talking about usability but about the quality and universality of the end result. word documents are overblown and render maths horribly. And the OP didn't even type anything that couldn't have been done in plain text here as can be seen by the fact that I literally cut and pasted the text into the thread window.

[matt grime]

what i have typed is:

lim a_n+1-a_n/b_n+1-b_n=G

 

yes that is what you typed now, not what you first wrote, and the tree's interpretation of either string of characters is strictly correct, it does equate to 1/1, because you have not braced off the subscripts. Sure, the correct meaning is clear if you stare at it ifor a while but you shouldn't have to do that.

[lqg]

yes that is what you typed now, not what you first wrote, and the tree's interpretation of either string of characters is strictly correct, it does equate to 1/1, because you have not braced off the subscripts. Sure, the correct meaning is clear if you stare at it ifor a while but you shouldn't have to do that.

sorry but i prefer to talk about the maths and not about the means to convey it via the internet.

 

i guess i should work it out by my own, as allways.

[the tree]

I have a hunch that there's something up with the question. If [math]

\lim_{n \to \infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=G[/math]

 

then [math]

\lim_{n \to \infty}(\frac{a_{n+1}}{b_{n+1}-b_n}+\frac{-a_n}{b_{n+1}-b_n})=G[/math]

 

so [math]\lim_{n \to \infty}(\frac{a_{n+1}}{-b_n}+\frac{-a_n}{-b_n})=G\lim_{n\to\infty}b_{n+1}[/math]

 

and[math]\lim_{n \to \infty}\frac{-a_n}{-b_n}=\lim_{n \to \infty}\frac{a_n}{b_n}=(G\lim_{n\to\infty}b_{n+1})-\frac{a_{n+1}}{-b_n}[/math]

 

see what I mean?

[lqg]

i'm not the expert, but that seems to me to be invalid.

[matt grime]

sorry but i prefer to talk about the maths and not about the means to convey it via the internet.

 

i guess i should work it out by my own' date=' as allways.[/quote']

 

 

if you do not express yourself accurately and concisely then you cannot claim to be 'talking about the maths'. if you want help and want to ask in a written forum like this you have the responsibility to make sure that what you write is unambiguous and correct; it is not up to the audience to attempt to interpret if, by a_n+1, you mean

 

[math]a_{n+1}[/math]

or

 

[math]a_n +1[/math]

 

if you want to communicate clearly take the effort to either learn the latex or typeset clearly using braces. the latter of these two options costs you zero time to do and greatly eases the time of those who voluntarily choose to help you.

 

dont' try and pretend it is up to others to figure out what you mean.

[matt grime]

I have a hunch that there's something up with the question. If [math]

\lim_{n \to \infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=G[/math]

 

then [math]

\lim_{n \to \infty}(\frac{a_{n+1}}{b_{n+1}-b_n}+\frac{-a_n}{b_{n+1}-b_n})=G[/math]

 

so [math]\lim_{n \to \infty}(\frac{a_{n+1}}{-b_n}+\frac{-a_n}{-b_n})=G\lim_{n\to\infty}b_{n+1}[/math]

 

and[math]\lim_{n \to \infty}\frac{-a_n}{-b_n}=\lim_{n \to \infty}\frac{a_n}{b_n}=(G\lim_{n\to\infty}b_{n+1})-\frac{a_{n+1}}{-b_n}[/math]

 

see what I mean?

 

 

no, not at all. the conclusion after the word 'so' is absolutely not valid;

[s pepperchin]

If you don't want to learn LaTeX the esiest way to import problems is to use PowerPoint you start with an empty slide, no textboxes or anything, and start by drawing a white box over most of the slide. This is so when you save it it will be visible. Next use the equation editor, to use it go to the menu and choose insert object and equation editor is one of the choices for object type. You can also place text boxes on the frame for writing out problems or you can draw diagrams. Once you have it the way you want use the arrow to draw a selection box outside of the white box. When you do this everything should be selected. You can then group everything together by right clicking on one of the selected items, and choosing to grouping from the menu. At this point you can right click on the group and select save as picture from the menu, then change the file type to JPEG the default isn't a valid file type in the forums. at this point you can place the picture in your post and by learning how to do this you will get practice making diagrams that you could use in reports. I am including an example of a file I've used before.

[the tree]

no, not at all. the conclusion after the word 'so' is absolutely not valid;

On closer inspection, that is complete bollocks. I'm not great with limits.

 

Still, the question asks for proof that [math]\lim_{n \to \infty}\frac{a_{n+1}-a_n}{b_{n+1}-b_n}=\lim_{n \to \infty}\frac{a_n}{b_n}[/math] wich just looks wrong to me.

 

If you don't want to learn LaTeX the esiest way to import problems is to use PowerPoint

But then you'd be using Powerpoint. What's wrong with plain text?

[matt grime]

Well, let's see if it is reasonable, suppose that b_n = n, and a_n=an+c for some choice of a and c, then the limit of the difference converges to a, and a_n/b_n also converges to a, so it seems perfectly reasonable to me. Note neither a_n or b_n converge.

 

Now suppose that b_n is some sequence converging to 1, say, and a_n=kb_n for some k, then the difference quotient converges to k, as does the quotient, and a_n and b_n converge, so it seems really quite reasonable, doesn't it?

 

Can we delete that bloody powerpoint slide that is big, a waste of badnwidth and off topic?

[s pepperchin]

its not about using powerpoint its about using the equation editor which allows you to make the equations look the way they do when you use LaTeX but equation editor gives you dropdown boxes to choose from so its more user friendly. Its a little extra work that makes it easier for others to understand exactly what problem your trying to solve.

[the tree]

As is typing clearly, and that doesn't require using PowerPoint.

[s pepperchin]

I bet that last post made you feel better about yourself, didn't it. Im just trying to offer another method for posting equations that makes them easy to read.