Prove using the definition of a limit, Please help

[silvermane]

Prove using the definition of a limit, Please help! :)

Homework Statement

Prove using only the definition of a limit, that the sequence:

\frac{n}{(n+1)^1/2} - \frac{n}{(n+2)^1/2} converges.

Homework Equations

Let E>0 and choose a special N = something*E that whenever n>N our difference of limits is less than E...

The Attempt at a Solution

I know that the limit is 0, but I'm having trouble finding the special N. The algebra for this is horrible and I've spent a long while working on it. Please help. It will be greatly appreciated.

 

[pellman]

hint #1
show that

\sqrt{\frac{n}{n+1}}-\sqrt{\frac{n}{n+2}}<\sqrt{\frac{n}{n+1}-\frac{n}{n+2}}

and use this fact to simplify what needs to be less than E. That is, find N such

\sqrt{\frac{n}{n+1}-\frac{n}{n+2}} is less than E for all n > N and you've got it.

 

[silvermane]

But I don't have a square root in the numerator; it's just in the denominator. I'm slightly confused, but will keep looking at it - in case it was me. :(

Here's what I did:
I have that my above sequence is equal to
\frac{3n^2}{n^2+3n+2} < \frac{3n^2}{n^2+3n} < E

and my N = \frac{2E}{3-E}

 

[pellman]

But I don't have a square root in the numerator; it's just in the denominator.

Sorry. i misread it.

I'm slightly confused, but will keep looking at it - in case it was me. :(

Here's what I did:
I have that my above sequence is equal to
\frac{3n^2}{n^2+3n+2} < \frac{3n^2}{n^2+3n} < E

and my N = \frac{2E}{3-E}

So you've got it now?

 

[silvermane]

Sorry. i misread it.



So you've got it now?

Well, I want to make sure that what I'm doing is correct. I've gone to my professor's office, and he wasn't very helpful to me. (there's a language barrier)

Either way, does my N make sense and is mathematically correct?
I wanted to get an N without squaring it as well, so I don't think what I've done above is allowed to be done.
Thank you so much for your help so far :)

 

[pellman]

Well, actually no. How did you get

\frac{3n^2}{n^2+3n+2}

?

 

[silvermane]

I squared everything and then simplified. I've come to realize however, that it's not something I shouldn't have done, but I wasn't thinking clearly at the time. I'm stuck when it comes down to the algebra: I know the limit is 0, so I just need to simplify my expression... I can then find a comparison to get a good N for my proof. I'm starting to become very frustrated with this problem. :(So far, I have this:

\frac{n}{(n+1)^1/2} - \frac{n}{(n+2)^1/2}

= \frac{(n^3 + n^2)^1/2 - (n^3 + 2n^2)^1/2}{(n^2 + 3n +2)^1/2}

Could I say that,

\frac{(n^3 + n^2)^1/2 - (n^3 + 2n^2)^1/2}{(n^2 + 3n +2)^1/2} < \frac{(n^3 + n^2)^1/2}{(n^2 + 3n +2)^1/2},

then

\frac{(n^3 + n^2)^1/2}{(n^2 + 3n +2)^1/2} < \frac{(n^3/2)}{(3n)^1/2} = \frac{n}{3^1/2}

But I think this won't work with the definition of a limit. Someone please help! Any hint would do, I don't want an answer. :(

 

[pellman]

lol. I don't think this converges. the numerator goes to infinity faster than the denominator. Numerator is ~n^1. Denominator is ~n^(1/2) .

 

[silvermane]

If it does diverge, could I show that using the definition of a limit and reach a contradiction? I've been working on this for days, and the way the question is worded, it leads the student to think that the series converges. This is just a problem to help me prepare for the final, since it is a problem in the practice final, and it models the true quite closely. If you think my math above is correct, then it must diverge under the definition of the limit.

But then I thought about it some more, and was wondering if this was true:

\frac{(n^3 + n^2)^(1/2) - (n^3 +2n^2)^(1/2)}{(n^2 + 3n + 2)^(1/2)} <

\frac{(n^2)^(1/2) - (2n^2)^(1/2)}{(n^2 + 3n)^(1/2)} = \frac{n(1-\sqrt{2})}{(n^2 + 3n)^(1/2)}

I'm really trying my best to work this out, but I can't see what I'm doing wrong here. If anyone could put me in the right direction, they would be a lifesaver. Thank you for helping me this much already pellman!

 

[pellman]

I think it is simpler than that. Let

a_n = \frac{n}{\sqrt{n+1}}-\frac{n}{\sqrt{n+2}}

It should be straightforward to show that a_{n+1} &gt; a_n for all n greater than some N. I bet N is rather low, probably 1 or 2. This will require some rather tedious algebra or some clever shortcuts, if there are any.

That's what I'd try first. It might be difficult though. I don't have time to work through it myself.

 

[╔(σ_σ)╝]

You are over complicating things.

<br /> a_n = \frac{n}{\sqrt{n+1}}-\frac{n}{\sqrt{n+2}}<br />

<br /> a_n = \frac{n \left( \sqrt {n+2} - \sqrt{n+1}\right)}{\sqrt{n+1} \sqrt{n+2}}<br />

Rationalize to get
<br /> a_n = \frac{n}{\sqrt{n+1} \sqrt{n+2}\left( \sqrt {n+2} + \sqrt{n+1}\right)}<br />

From here it is straight forward to show that
<br /> a_n &lt; \frac{1}{2 \sqrt{n}}<br />

So it is easy to show the limit is zero.

 

[pellman]

Sheesh. I'm getting old.

 

[silvermane]

You are over complicating things.

<br /> a_n = \frac{n}{\sqrt{n+1}}-\frac{n}{\sqrt{n+2}}<br />

<br /> a_n = \frac{n \left( \sqrt {n+2} - \sqrt{n+1}\right)}{\sqrt{n+1} \sqrt{n+2}}<br />

Rationalize to get
<br /> a_n = \frac{n}{\sqrt{n+1} \sqrt{n+2}\left( \sqrt {n+2} + \sqrt{n+1}\right)}<br />

From here it is straight forward to show that
<br /> a_n &lt; \frac{1}{2 \sqrt{n}}<br />

So it is easy to show the limit is zero.

lol I actually did something like that, and worked it out last night. I ended up getting 1/E^2 :)

Combining and THEN taking the conjugate was what needed to be done. Thank you so much for your help! I feel very prepared for my final now

 

[silvermane]

PS: you're not old, you're awesome :)

Heck, I couldn't figure it out, and MANY others I went to didn't even know where to start. I'm very happy that I was able to finally get it, and I think you should be too. :)
You deserve a pat on the back!