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Math Help - Struggling to prove convergene using definition of convergence

  1. #1
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    Struggling to prove convergene using definition of convergence

    Hi, first post, found myself stuttering a bit so just trying to get my head round some examples of proving certain sequences converge.

    The first one, I think I did correctly, prove that

    (an) = (5n^2+srt(n))/n^2 converges to five

    Obviously you have to show that the modulus of an-l is less than episilon for all n.
    I found myself rearranging (an)-l and got (an-l)=n^-(3/2).
    So then I took epsilon to be an arbitary real number, greater than sqrt(1/N^3)
    Can someone confirm that please?


    The second one, I'm struggling to even put it into a suitable format for an-l, it is show that the sequence:

    (an)= (6+5n)/(2-3n) converges to -5/3.
    It's probably pretty basic algebra, but I'm struggling to even get this into the form where I can pick an arbitrary epsilon and take an N.
    So could do with some help on this one.

    Apologies if the typing looks a bit ungainly, I'll try my best to fit in here
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  2. #2
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    Quote Originally Posted by MickHarford View Post
    The first one, I think I did correctly, prove that

    (an) = (5n^2+srt(n))/n^2 converges to five
    Obviously you have to show that the modulus of an-l is less than episilon for all n.
    I found myself rearranging (an)-l and got (an-l)=n^-(3/2).
    So then I took epsilon to be an arbitary real number, greater than sqrt(1/N^3)
    Can someone confirm that please?
    Actually we start with [b]any[/u] \varepsilon  > 0.
    Then we work from there.
    We can see that  \dfrac{1}{{\sqrt[3]{{\varepsilon ^2 }}}} > 0 so
    \left( {\exists N} \right)\left[ {n \geqslant N\, \Rightarrow \,n \geqslant \dfrac{1}{{\sqrt[3]{{\varepsilon ^2 }}}}} \right] .
    Then we see \dfrac{1}{{\sqrt {n^3 } }} < \varepsilon
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