For the sequence a_n = n^-3 and e>0 find N(e) such that for all n>=N(e), |a_n| < e?

January 9, 2009 on 4:12 pm | By | In bluefreesky.com | For the sequence a_n = n^-3 and e>0 find N(e) such that for all n>=N(e), |a_n| < e?
  • n belongs to natural numbers, N(e) belongs to natural numbers. e is epsilon and a_n is a subscript n


  • Pick N such that N > 1/∛ε

    Now, for all n > N we have n > 1/∛ε which means

    1/n < ∛ε or

    1/n³ < ε for all n > N.

    HTH

    Charles


  • So we have that,

    a_n = 1/n³

    Working backwards, suppose that we have that,

    a_n = 1/n³ < ε

    Therefore, we have that

    1/n³ < ε
    n³ > 1/ε;
    n > 1/∛ε

    Therefore, the smallest N that will work is N = 1/∛ε.

    Hope this helps!

    Cautionary Note: When writing up a proof for this, don't show the work I did. Simply state it like Charles does, but if you were wondering how Charles came up with N then that's where my answer comes in handy.







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