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    I have been working on this question:

    Express the finite series

     \frac{1}{\log_2 a} + \frac{1}{\log_3 a}+...+\frac{1}{\log_n a}

    as a quotient of logarithms to base 2. Does the series converge as n \to \infty ?

    I changed the base of each of the logarithms in the denominators to 2. This left me with the expression

     \frac{\log_2 2 + \log_2 3 + ...+ \log_2 n }{\log_2 a} = \frac{\log_2 (2 \cdot 3 \cdot 4... (n-1) \cdot n)}{\log _2 a}.

    I think that as  n \to \infty the quotient will grow without bound as log is an increasing function. However I am not sure whether this is correct...
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    (Original post by FXLander)
    I have been working on this question:

    Express the finite series

     \frac{1}{\log_2 a} + \frac{1}{\log_3 a}+...+\frac{1}{\log_n a}

    as a quotient of logarithms to base 2. Does the series converge as n \to \infty ?

    I changed the base of each of the logarithms in the denominators to 2. This left me with the expression

     \frac{\log_2 2 + \log_2 3 + ...+ \log_2 n }{\log_2 a} = \frac{\log_2 (2 \cdot 3 \cdot 4... (n-1) \cdot n)}{\log _2 a}.

    I think that as  n \to \infty the quotient will grow without bound as log is an increasing function. However I am not sure whether this is correct...
    Could also be written as \dfrac{\log_2(n!)}{\log_2a}

    And yes, numerator is unbounded, denominator is a constant, hence off to infinity we go.
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    (Original post by FXLander)
    I have been working on this question:

    Express the finite series

     \frac{1}{\log_2 a} + \frac{1}{\log_3 a}+...+\frac{1}{\log_n a}

    as a quotient of logarithms to base 2. Does the series converge as n \to \infty ?

    I changed the base of each of the logarithms in the denominators to 2. This left me with the expression

     \frac{\log_2 2 + \log_2 3 + ...+ \log_2 n }{\log_2 a} = \frac{\log_2 (2 \cdot 3 \cdot 4... (n-1) \cdot n)}{\log _2 a}.

    I think that as  n \to \infty the quotient will grow without bound as log is an increasing function. However I am not sure whether this is correct...
    Note that you can see that this diverges directly: suppose that n > a and that we want to find x=\log_n a \Rightarrow n^x=a. It is clear(*) that 0 < x < 1 so that \frac{1}{x}=\frac{1}{\log_n a} >1 so once n>a, we are adding a series all of whose terms are bigger than 1. So it diverges.

    (*) If it's not clear, use specific numbers e.g:

    n=5,a=5 \Rightarrow \log_5 5 =1 \\ n=10,a=5 \Rightarrow \log_{10} 5 < 1 \\ n=100,a=5 \Rightarrow \log_{100} 5 < 1, \cdots
 
 
 
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