Suppose that {s_n} is a sequence of positive numbers.?

Prove that if lim (s_n)^2 = s^2 then lim (s_n) = s

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  • kb
    Lv 7
    8 years ago
    Favorite Answer

    Since {s(n)} is a sequence of positive numbers, its limit can not be negative (ruling out -s, assuming that s is non-negative).

    We know that {(s(n))^2} is bounded (as it converges):

    So, 0 < (s(n))^2 < M for some M > 0 (and for all n)

    ==> 0 < s(n) < √M for all n.

    Now, let ε > 0 be given.

    Since {(s(n))^2} converges to s^2, there exists a positive integer N such that

    |(s(n))^2 - s^2| < ε/(s + √M) for all n > N.

    So, |(s(n))^2 - s^2|

    = |s(n) - s| * (s(n) + s)

    < |s(n) - s| * ε/(s + √M)

    = ε for all n > N.

    Hence, {s(n)} converges to s.

    I hope this helps!

  • 8 years ago

    This is a particular case of the more general property

    Let f be any continuous function,

    if an → a then f(an) → f(a)

    in your question f(x) = √x

    http://en.wikipedia.org/wiki/Limit_of_a_sequence#P...

    ===================================================

    if (Sn)² → S² then Sn → S

    suppose S > 0

    let ε> 0 be any positive real number such that

    ε < S.............(*)

    ε² < S², there exists an index N such that

    if n>N then |S² - (Sn)²| < ε²

    S² - ε² < (Sn)² < S² + ε²

    S² + ε² < (S + ε)² = S² + ε² + 2εS

    therefore

    √(S² + ε²) < S + ε,

    for positive S and ε

    √(S² - ε²) > S - ε

    square both sides

    S² - ε² > S² + ε² - 2εS

    - 2 ε² > - 2εS

    ε < S, which is true for (*)

    from

    S² - ε² < (Sn)² < S² + ε²

    taking square root of all terms

    S - ε < √(S² - ε²) < Sn < √(S² + ε²) < S + ε,

    S - ε < Sn < S + ε,

    |S - Sn| < ε

    QDE

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