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Math Help - Limit of product of sequences

  1. #1
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    Limit of product of sequences

    Evaluate lim sqrt(n)*[sqrt(n+1)-sqrt(n)]

    I know that limit sqrt(n) = Infinity and that limit (sqrt(n+1)-sqrt(n)) = 0. And I know that sqrt(n)*(sqrt(n+1)-sqrt(n)) = sqrt(n)/(sqrt(n+1)+sqrt(n)).

    I believe the limit of the product of these sequences is 1/2, but I am not sure how to get there. Thanks for your help.
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  2. #2
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    This might be very simplistic, but could you use l'Hospital's Rule? It would certainly work for the real variable case, and would give you, as you intimated, 1/2.
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  3. #3
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    Quote Originally Posted by tarheelborn View Post
    Evaluate lim sqrt(n)*[sqrt(n+1)-sqrt(n)]
    I know that sqrt(n)*(sqrt(n+1)-sqrt(n)) = sqrt(n)/(sqrt(n+1)+sqrt(n)).
    I believe the limit of the product of these sequences is 1/2, but I am not sure how to get there. Thanks for your help.
    \frac{{\sqrt n }}{{\sqrt {n + 1}  + \sqrt n }} = \frac{1}<br />
{{\sqrt {1 + \frac{1}{n}}  + 1}}
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  4. #4
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    Thank you for the simplification. Now I need to find N and I am not sure how to combine sqrt[(1/n)+1]+sqrt(n). Or would I be able to ignore the +1 part of it and somehow combine sqrt(1/n) and sqrt(n)?
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  5. #5
    Member mohammadfawaz's Avatar
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    why do u need to find N? you're asked to find the limit. Just find it! u know \lim_{n->\infty}\frac{1}{n}=0 then you're done be applying this to Plato's note
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  6. #6
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    I have to prove the limit, so I definitely do need to find N. Thanks.
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  7. #7
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    In that case, go back to Plato's formula: and try to find N such that for a given \epsilon and if n>N, then, |\frac{1}{\sqrt{1+\frac{1}{n}}+1} - \frac{1}{2}|<\epsilon
    i.e. \frac{\sqrt{1+\frac{1}{n}}-1}{2\sqrt{1+\frac{1}{n}}+2}<\epsilon (note that we remove the absolute value signs - this quantity is positive!).
    Now, this means: \frac{\sqrt{1+\frac{1}{n}}-1}{2\sqrt{1+\frac{1}{n}}+2}<\frac{\sqrt{1+\frac{1}  {n}}-1}{4}<\epsilon for large n (check this!)
    Therefore: \sqrt{1+\frac{1}{n}}-1<4\epsilon
    Hence: \sqrt{1+\frac{1}{n}}<1+4\epsilon
    1+\frac{1}{n}<1+16\epsilon^2+8\epsilon
    n>\frac{1}{16\epsilon^2+8\epsilon} You found your N!
    Note that you don't have to find the best value of N. Any value that works if fine to prove the limit.
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  8. #8
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    Thank you so much!
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  9. #9
    Member mohammadfawaz's Avatar
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    Just one more thing to not: N is an integer! (obviously)...So, N is equal to the smallest integer larger than \frac{1}{16\epsilon^2+8\epsilon}.
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