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Math Help - divisor convergence?

  1. #1
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    divisor convergence?

    If a_n > 0 and (an+1/ a_n) converges to p and p<1, prove a_n converges to zero.
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cowgirl123 View Post
    If a_n > 0 and (an+1/ a_n) converges to p and p<1, prove a_n converges to zero.
    consider this:

    If \frac {a_{n + 1}}{a_n} \to p < 1, it must mean that a_n >a_{n + 1}. But this means that \{ a_n \} is strictly decreasing. But a_n > 0 for all n. Which means that we are bounded below by 0. Thus a_n is a monotonically decreasing sequence that is bounded below (by zero). Now say that more formally
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  3. #3
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    Quote Originally Posted by cowgirl123 View Post
    If a_n > 0 and (an+1/ a_n) converges to p and p<1, prove a_n converges to zero.
    This means, \lim \frac{a_{n+1}}{a_n} = p. It must be the case that p\geq 0 because the sequence of ratios is a sequence of non-negative terms. So the limit is non-negative. I leave the case p=0 to you to prove. I will assume p>0, i.e. 0<p<1. The important step is to notice that there most exists \epsilon > 0 so that p+\epsilon < 1. This means for \epsilon >0 we have by convergence, \left| \frac{a_{n+1}}{a_n} - p \right| < \epsilon for n\geq N\in \mathbb{N}. Thus, \frac{a_{n+1}}{a_n} - p < \epsilon \implies a_{n+1} < (p+\epsilon)a_n for n\geq N.
    This means,
    a_{N+1} < (p+\epsilon)a_N
    a_{N+2} < (p+\epsilon)a_N < (p+\epsilon)^2 a_N
    a_{N+3} < (p+\epsilon)a_{N+2} < (p+\epsilon)^2 a_{N+1} < (p+\epsilon )^3 a_N.
    In general,
    a_{N+k} < (p+\epsilon)^k a_N for k\geq 1.
    This tells us that the sequence \{ a_ n \} is bounded by (p+\epsilon)^{n-N} a_N for all n\geq N.
    Thus,
     0 < a_n < (p+\epsilon)^n a_n (p+\epsilon)^{-N}.
    Now, \lim \ (p+\epsilon)^n a_n (P+\epsilon)^{-N} = 0.
    Because 0 < (p+\epsilon) < 1 and it is a geometric sequence.
    Q.E.D.
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