Prove using the definition of a limit, Please help! :)by silvermane Tags: convergence, definition of limit, epsilon, limit 

#1
Dec710, 11:54 AM

PF Gold
P: 117

1. The problem statement, all variables and given/known data
Prove using only the definition of a limit, that the sequence: [tex]\frac{n}{(n+1)^1/2}[/tex]  [tex]\frac{n}{(n+2)^1/2}[/tex] converges. 2. Relevant equations Let E>0 and choose a special N = something*E that whenever n>N our difference of limits is less than E... 3. 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. 



#2
Dec710, 12:27 PM

P: 565

hint #1
show that [tex]\sqrt{\frac{n}{n+1}}\sqrt{\frac{n}{n+2}}<\sqrt{\frac{n}{n+1}\frac{n}{n+2}}[/tex] and use this fact to simplify what needs to be less than E. That is, find N such [tex]\sqrt{\frac{n}{n+1}\frac{n}{n+2}}[/tex] is less than E for all n > N and you've got it. 



#3
Dec710, 12:38 PM

PF Gold
P: 117

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 [tex]\frac{3n^2}{n^2+3n+2}[/tex] < [tex]\frac{3n^2}{n^2+3n}[/tex] < E and my N = [tex]\frac{2E}{3E}[/tex] 



#4
Dec710, 03:28 PM

P: 565

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



#5
Dec810, 03:34 PM

PF Gold
P: 117

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 :) 



#6
Dec810, 04:15 PM

P: 565

Well, actually no. How did you get
[tex]\frac{3n^2}{n^2+3n+2}[/tex] ? 



#7
Dec810, 04:57 PM

PF Gold
P: 117

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: [tex]\frac{n}{(n+1)^1/2}[/tex]  [tex]\frac{n}{(n+2)^1/2}[/tex] = [tex]\frac{(n^3 + n^2)^1/2  (n^3 + 2n^2)^1/2}{(n^2 + 3n +2)^1/2}[/tex] Could I say that, [tex]\frac{(n^3 + n^2)^1/2  (n^3 + 2n^2)^1/2}{(n^2 + 3n +2)^1/2}[/tex] < [tex]\frac{(n^3 + n^2)^1/2}{(n^2 + 3n +2)^1/2}[/tex], then [tex]\frac{(n^3 + n^2)^1/2}{(n^2 + 3n +2)^1/2}[/tex] < [tex]\frac{(n^3/2)}{(3n)^1/2}[/tex] = [tex]\frac{n}{3^1/2}[/tex] 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. :( 



#8
Dec910, 07:57 AM

P: 565

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) .




#9
Dec910, 09:19 AM

PF Gold
P: 117

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: [tex]\frac{(n^3 + n^2)^(1/2)  (n^3 +2n^2)^(1/2)}{(n^2 + 3n + 2)^(1/2)}[/tex] < [tex]\frac{(n^2)^(1/2)  (2n^2)^(1/2)}{(n^2 + 3n)^(1/2)}[/tex] = [tex]\frac{n(1\sqrt{2})}{(n^2 + 3n)^(1/2)}[/tex] 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! 



#10
Dec910, 12:19 PM

P: 565

I think it is simpler than that. Let
[tex]a_n = \frac{n}{\sqrt{n+1}}\frac{n}{\sqrt{n+2}}[/tex] It should be straightforward to show that [tex]a_{n+1} > a_n[/tex] 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. 



#11
Dec910, 10:15 PM

P: 851

You are over complicating things.
[tex] a_n = \frac{n}{\sqrt{n+1}}\frac{n}{\sqrt{n+2}} [/tex] [tex] a_n = \frac{n \left( \sqrt {n+2}  \sqrt{n+1}\right)}{\sqrt{n+1} \sqrt{n+2}} [/tex] Rationalize to get [tex] a_n = \frac{n}{\sqrt{n+1} \sqrt{n+2}\left( \sqrt {n+2} + \sqrt{n+1}\right)} [/tex] From here it is straight forward to show that [tex] a_n < \frac{1}{2 \sqrt{n}} [/tex] So it is easy to show the limit is zero. 



#12
Dec1010, 06:28 AM

P: 565

Sheesh. I'm getting old.




#13
Dec1010, 08:29 AM

PF Gold
P: 117

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 


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