Hey there! Sign in to join this conversationNew here? Join for free
    • Thread Starter
    Offline

    9
    ReputationRep:
    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...
    • Study Helper
    Online

    13
    (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.
    Offline

    11
    ReputationRep:
    (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
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Did TEF Bronze Award affect your UCAS choices?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.