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Limit proof

  1. Sep 5, 2015 #1
    Edit: the red \E is the epsilon sign

    1. The problem statement, all variables and given/known data

    ##Given\quad { x }_{ n }=\sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \\ Prove\quad that:\quad \lim _{ n\rightarrow \infty }{ { x }_{ n } } =0##

    2. Relevant equations

    ##For\quad any\quad arbitrary\quad \E >0,\quad there\quad exists\quad an\quad N>0\quad such\quad that\\ if\quad n>N\quad \Longrightarrow \quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| <\E ##


    3. The attempt at a solution

    ##\left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| \quad \le \quad \sqrt { n+1 } +\sqrt { n+2 } +2\sqrt { n+3 } \\ \\ \\ <\quad \sqrt { n+3 } +\sqrt { n+3 } +2\sqrt { n+3 } \quad =\quad 4\sqrt { n+3 } \quad \\ \\ <\quad \E \\ \\ Taking\quad N\quad =\quad -3\quad +\quad \left( \frac { \E }{ 4 } \right) ^{ 2 }\\ if\quad n>N\quad \Longrightarrow \quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| <\quad \E \\ thus\quad by\quad the\quad definition\quad of\quad a\quad limit,\quad the\quad limit\quad is\quad 0.##

    However, whenever I take any epsilon, say 0.1, I get N=-2.999375, so if I choose n = 1 for example, and I sub it into the inequality, I get 0.85337.... < 0.1 which is obviously incorrect. Any idea what part is wrong?
     
    Last edited: Sep 5, 2015
  2. jcsd
  3. Sep 5, 2015 #2

    PeroK

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    You've effectively tried to show that ##4\sqrt{n+3}## tends to ##0## an ##n## tends to ##\infty##. You've done this by taking ##n## as a specific negative number. Which is absurd.
     
  4. Sep 5, 2015 #3
    I am not sure if I understand you. can you elaborate just a bit more if you don't mind? Also, what other inequalities can i use to solve this?
     
  5. Sep 5, 2015 #4

    PeroK

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    You started by showing that ##|x_n| < 4\sqrt{n+3}##. But this is useless if you are trying to show that ##x_n \rightarrow 0##. Can you see that?
     
  6. Sep 5, 2015 #5
    so you mean that I need to use the inequalities correctly to reach a function that is bigger than ##|x_n| and at the same time tends towards zero?
     
  7. Sep 5, 2015 #6

    PeroK

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    Yes!
     
  8. Sep 5, 2015 #7

    PeroK

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    Here's a hint. Try to show that ##\sqrt{n+1} - \sqrt{n+3} \rightarrow 0## as a first step.
     
  9. Sep 5, 2015 #8
    Ok so using yout hint, I reached this:

    ##\left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| =2\sqrt { n+3 } -\quad \sqrt { n+1 } -\sqrt { n+2 } \quad =\quad \frac { 3n+10\quad -\quad \sqrt { n+1 } \left( 2\sqrt { n+3 } \cdot \sqrt { n+2 } \right) }{ 2\sqrt { n+3 } \cdot \sqrt { n+2 } } \quad <\quad \frac { 3n+10\quad -\quad \sqrt { n } \left( 2\sqrt { n } +\sqrt { n } \right) }{ 2\sqrt { n+3 } \cdot \sqrt { n+2 } } \\ <\quad \frac { 10 }{ \sqrt { n+2 } } .\quad Taking\quad N\quad =\quad \left( \frac { 10 }{ \E } \right) ^{ 2 }-2,\quad then\quad n>N\quad implies\quad \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| <\E ##

    Is there any simpler way that I've missed?
    Also, thanks for your help.
     
  10. Sep 5, 2015 #9

    PeroK

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    I'm not sure how that uses my hint! How did you get that equality where ##3n + 10## appears? It doesn't look right in any case.
     
    Last edited: Sep 5, 2015
  11. Sep 5, 2015 #10
    ##Using\quad your\quad hint:\\ \left| \sqrt { n+1 } -\sqrt { n+3 } \right| =\sqrt { n+3 } -\sqrt { n+1 } =\quad \frac { n+3-n-1 }{ \sqrt { n+1 } +\sqrt { n+3 } } <\frac { 2 }{ \sqrt { n+1 } +\sqrt { n+3 } } <\frac { 2 }{ \sqrt { n } } ,\quad therefore\quad take\quad N=\left( \frac { 2 }{ \E } \right) ^{ 2 }\\ Using\quad the\quad same\quad method\quad but\quad with\quad the\quad whole\quad thing:\\ \left| \sqrt { n+1 } +\sqrt { n+2 } -2\sqrt { n+3 } \right| =2\sqrt { n+3 } -\quad \sqrt { n+2 } -\sqrt { n+1 } \quad =\quad \left( 2\sqrt { n+3 } -\quad \sqrt { n+2 } -\sqrt { n+1 } \right) \cdot \frac { 2\sqrt { n+3 } +\quad \sqrt { n+2 } }{ 2\sqrt { n+3 } +\quad \sqrt { n+2 } } =\frac { 4(n+3)-(n+2)-\sqrt { n+1 } (2\sqrt { n+3 } +\quad \sqrt { n+2 } ) }{ 2\sqrt { n+3 } +\quad \sqrt { n+2 } } =\frac { 3n+10-\sqrt { n+1 } (2\sqrt { n+3 } +\quad \sqrt { n+2 } ) }{ 2\sqrt { n+3 } +\quad \sqrt { n+2 } } \\ ##

    I just feel that there has to be someway easier but I just can't get it.Could you make it easier?
     
  12. Sep 5, 2015 #11

    PeroK

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    If you have two sequences that tend to 0, then what can you say about the sum of the two?
     
  13. Sep 5, 2015 #12
    Well if both tend towards 0 then using the arithmetic rule, then the sum of them tends towards 0. Do you mean that I should divide it to two such as root(n+3) - root(n+1) once and show it tends towards 0 then use do root(n+3) - root(n+2) and show that this also tends towards 0 and using the arithmetic law of limits, this shows that the sum of these two sequences converge to 0 as a result? Am I allowed to use the Limit laws in these proofs as facts without proving them? I thought i can only use algebraic manipulation?

    Also, one last question, would the method I used be considered incorrect? The one for the 3 of them simultaneously?
     
  14. Sep 5, 2015 #13

    PeroK

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    If you can't use the arithmetic rule generally, you can always use it specifically:

    If ##|y_n| < \frac{1}{\sqrt{n}}## and ##|z_n| < \frac{1}{2\sqrt{n}}## then for ##|x_n| = |y_n + z_n| \dots##
     
  15. Sep 5, 2015 #14
    Aha, I see.
    Also, would the way I used with lots of algebra be considered a correct proof or faulty?
     
  16. Sep 5, 2015 #15

    PeroK

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    It's not wrong, but I would never have thought of doing it that way. The first thing I thought of was to split it into two. And, in fact, I calculated that:

    ##|\sqrt{n+a} - \sqrt{n + b}| < |\frac{a-b}{2\sqrt{n}}|## when ##a \ne b##

    Which, I feel, is a better way to think about mathematics. There's nothing special about the numbers 1, 2 and 3 in this case.
     
  17. Sep 5, 2015 #16
    1
    Thanks a lot for your help, and sorry if I bothered you with my not so intelligent questions :)
     
  18. Sep 5, 2015 #17

    Ray Vickson

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    Your first line is wrong; actually, it make a correct statement, but is useless for what you want to prove. Of course the inequality
    [tex] | \,\sqrt{n+1} + \sqrt{n+2} - 2 \sqrt{n+3}\,| \leq \sqrt{n+1} + \sqrt{n+2} + 2 \sqrt{n+3} [/tex]
    is perfectly true, but since the right-hand-side ##\to \infty## as ##n \to \infty##, that does not help you to establish that the left-hand-side becomes smaller and smaller as ##n## increases towards ##\infty##.
     
  19. Sep 6, 2015 #18
    Notice that you have yn, yn+1 and yn+2, such that xn = yn + yn+1 - 2yn+2 combine this with the fact that finitie elements do not alter the limit of a sequence and you should reach the desired limit.

    Edit: i am wrong here, the problem is that my statement is not true unless yn converges, and here it doesnt.
     
    Last edited: Sep 6, 2015
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