an=sum k=0 to k=n from 10 (-k^2).The limit of this sequence belongs to the set a)Q b) [-1,0) c) R/Q?

Update:

is 10 ^(-k^2)

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  • kb
    Lv 7
    7 years ago
    Favorite Answer

    a(n) = Σ(k = 0 to ∞) 10^(-k^2) is irrational (choice C, assuming you mean R\Q).

    -----------------

    We will actually prove a bit more, showing that it is transcendental.

    Let p(n) = Σ(k = 0 to n) 10^(n - k^2) and q(n) = 10^n.

    Then, |Σ(k = 0 to ∞) 10^(-k^2) - p(n)/q(n)|

    = Σ(k = n+1 to ∞) 10^(-k^2)

    = 10^(-(n+1)^2) + 10^(-(n+2)^2) + 10^(-(n+3)^2) + ...

    = 10^(-(n+1)^2) [1 + 10^(-(2n+3)) + 10^(-(4n+8)) + ...]

    < 10^(-(n+1)^2) [1 + 10^(-2n) + 10^(-4n) + ...]

    = 10^(-(n+1)^2) * (1/(1 - 10^(-2))

    = 10^(-(n+1)^2) * 100/99

    < 10^(-n^2) for sufficiently large n

    = 1/(q(n))^n.

    Hence, a(n) is a Liouville number and thus transcendental (and irrational).

    Link:

    http://en.wikipedia.org/wiki/Liouville_numbers

    I hope this helps!

  • 7 years ago

    well, the series = 10 + 1/10 + 1 / 10^4 + ... > 10, so it is not (b).

    you never said what Q or R was though. Unless you mean just Q = rationals, R = reals, but then I do not know what R / Q means. I know what R \ Q means, but not what is written in (c)

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