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Competition problems: 1. sequences/convergence 2. matrices

  1. Dec 6, 2011 #1
    1. The problem statement, all variables and given/known data
    I don't remember the exact problems but I'll try to recall it as best as I can.

    Given two positive real sequences a[itex]_{n}[/itex], b[itex]_{n}[/itex], with a[itex]_{1}[/itex] = b[itex]_{1}[/itex] = 1, and b[itex]_{n}[/itex] = b[itex]_{n-1}[/itex]a[itex]_{n}[/itex] - 2. Show that [itex]\sum^{\infty}_{n=2}[/itex] [itex]\frac{1}{a_{1}a_{2}\ldotsa_{n}}[/itex] converges and find what it converges to.

    3. The attempt at a solution

    To show that it converges, I tried to show that all a[itex]_{i}[/itex] from 2 to infinity have to be greater than 1. In other words, I want to show that

    b[itex]_{n}[/itex] = b[itex]_{n-1}[/itex]a[itex]_{n}[/itex] - 2 > b[itex]_{n-1}[/itex] - 2.

    So, I tried to show that by induction. First I had to find a[itex]_{2}[/itex], and found using the original sequence inequality that a[itex]_{2}[/itex] > 2 since all the b[itex]_{n}[/itex]'s are positive. Then I lost myself somewhere and just twiddled my thumbs for about 2 hours.

    [a]1. This is going to be badly worded cause I'm using memory recall but: For an n x n matrix with integer values, find n for the matrix such that when you dot product a row vector to itself, you get an even number and when you multiply it to any other row vector in that matrix, you get an odd number.

    3. I tried to find it by calling that initial matrix A and multiplying it to itself but I think I should have multiplied it to A[itex]^{t}[/itex] and got a matrix B with even numbers along the diagonal and odd numbers off the diagonal (to fit the two properties given), then tried to think of a way to find n using that matrix B. In other words, I had a staring contest with a blank piece of paper.

    Anyone have any ideas/solutions, mostly interested in the how to tackle such problems and show a proper proof for them and why they sell hot dogs in packages of 10 and buns in packages of 8?
  2. jcsd
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