Math

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If there is a recursively defined sequence such that

a1 = sqrt(2)
an + 1 = sqrt(2 + an)

Show that an < 2 for all n ≥ 1

  • Math -

    a1 = √2
    a2 = √(2 + √2)
    a3 = √(2 + √(2 + √2))
    I see a pattern

    an = √( 2 + √(2 + ...

    let x = √( 2 + √(2 + ...
    square both sides, that will drop the left-most √ on the right side

    x^2 = 2 + √( 2 + √(2 + ...
    x^2 - 2 = √( 2 + √(2 + ...
    but the right side is what I originally defined as x
    so
    x^2 - 2 = x
    x^2 - x - 2 = 0
    (x-2)(x+1) = 0
    x = 2 or x=-1
    (clearly x = -1 is an extraneous solution)

    so an must have a limit of 2.
    and since n is a finite number, term an < 2 .

    an interesting loop on the calculator is this

    1) enter √2
    2) press =
    3) plus 2
    4) press =
    5) press √
    6) repeat step 2)

    You should get appr 2 correct to 5 decimals after about 10 iterations.

  • Math -

    You prove that the sequences converges to 2, but you don't necessarily prove that it will never exceed 2...

    Thanks for the great help though

  • Math -

    I proved that its limit is 2
    i.e. it will actually never reach 2, only if I take all of its infinite terms.
    so clearly it can never exceed 2 if it can't reach 2.

  • Math -

    You are absolutely correct. Thanks so much for the help on this!

    I'd love to pay you guys back for this service :)

  • Math -

    The only thing missing here is that you must show that the series is monotonously increasing, i.e. that
    a_{n+1} > a_{n} (that should be easy). This, combined with the fact that the limit is 2 proves that a_n < 2.

    If you have a sequence that is not monotonously increasing (or decreasing), it isn't necesarily true that the limiting value isn't attained for finite n. Take e.g. the sequence

    sin(pi n/7)/n

    which tends to zero, but it is also equal to zero for n a multiple of 7.

  • Math -

    Brianna arrived at county fair at 10:25. she left at 11:40. how long was she at the fair?

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