Turn on thread page Beta
    • Thread Starter
    Offline

    0
    ReputationRep:
    ..
    Offline

    0
    ReputationRep:
    (Original post by thomas575)
    Sorry tried to write in latex but wouldn't let me, error apparently....
    it is to prove by induction the sum of 1/k from 1 to 2^(n-1) is greater than or equal to (n+1)/2

    i was told to write down the sums for n=1,2,3,4,5
    so....for
    n=1: 1/1
    n=2: 1/1+1/2
    n=3: 1/1+1/2+1/3+1/4
    n=4: 1/1+1/2+1/3+...+1/8
    n=5: 1/1+1/2+1/3+................+1/16
    Obviosuly i can see each time there is double the amount of fractions but cannot see how this is going to help me prove the above.
    With induction, you do the following:

    1) Show rule to be true for base case, n = 1.
    2) Assume the rule holds for n = k.
    3) Using 2), prove that the rule holds for n = k+1.

    Can you do it using this procedure?
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Phugoid)
    With induction, you do the following:

    1) Show rule to be true for base case, n = 1.
    2) Assume the rule holds for n = k.
    3) Using 2), prove that the rule holds for n = k+1.

    Can you do it using this procedure?
    sorry should have mentioned. I know the procedure for induction, but it's that inductive step i can't seem to figure out where to start showing that it holds for k+1 once i've assumed that P(k) is true
    Offline

    11
    ReputationRep:
    (Original post by thomas575)
    Sorry tried to write in latex but wouldn't let me, error apparently....
    it is to prove by induction the sum of 1/k from 1 to 2^(n-1) is greater than or equal to (n+1)/2

    i was told to write down the sums for n=1,2,3,4,5
    so....for
    n=1: 1/1
    n=2: 1/1+1/2
    n=3: 1/1+1/2+1/3+1/4
    n=4: 1/1+1/2+1/3+...+1/8
    n=5: 1/1+1/2+1/3+................+1/16
    Obviosuly i can see each time there is double the amount of fractions but cannot see how this is going to help me prove the above.
    Re-write the above statement as a summation.

    So, P(n)=\displaystyle \sum_{k=1} ^{2^{n-1}} \frac{1}{k}= \frac{n+1}{2}

    Now you want to show this holds for n+1. So add on the extra term, and then assume the above statement P(n) is true.
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mathematician!)
    Re-write the above statement as a summation.

    So, P(n)=\displaystyle \sum_{k=1} ^{2^{n-1}} \frac{1}{k}= \frac{n+1}{2}

    Now you want to show this holds for n= m+1. So add on the extra term, and then assume the above statement P(n) is true.
    will it make alot of difference that it's an inequality though?
    Offline

    11
    ReputationRep:
    (Original post by thomas575)
    will it make alot of difference that it's an inequality though?
    Sorry, forgot to write it as the inequality. But it makes no difference to just use the equality statement (greater than or equal to is the same as = or >, and proving just one of these statements should be sufficient.)
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by Mathematician!)
    Sorry, forgot to write it as the inequality. But it makes no difference to just use the equality statement (greater than or equal to is the same as = or >, and proving just one of these statements should be sufficient.)
    Also sorry i know i'm being a complete nuisance....!
    The next term will not just be 1 term though because of the upper limit in the sigma notation is 2^n-1 not n. So as shown by writing down the sums it will be well i don't know how many more terms, can't seem to think atm. So how would i write this...?
    Offline

    11
    ReputationRep:
    (Original post by thomas575)
    Also sorry i know i'm being a complete nuisance....!
    The next term will not just be 1 term though because of the upper limit in the sigma notation is 2^n-1 not n. So as shown by writing down the sums it will be well i don't know how many more terms, can't seem to think atm. So how would i write this...?
    It's 2^(n-1) to get it into the set of integers (I think).
    Anyway, you need to add on the extra term 1/2^n to the summation during the inductive step.
    Offline

    18
    ReputationRep:
    When you go from n to n+1, how many extra terms are there in the sum?
    And what is the smallest extra term?
    Then the sum of the extra terms >= "number of extra terms" x "smallest extra term"
    • Thread Starter
    Offline

    0
    ReputationRep:
    (Original post by DFranklin)
    When you go from n to n+1, how many extra terms are there in the sum?
    And what is the smallest extra term?
    Then the sum of the extra terms >= "number of extra terms" x "smallest extra term"
    Thanks, think this may get me there (i really hope so, sick of this lol). is it fair to say then that for the sum as above from 1 to 2^n then, when i do the inductive step i should get that it is greater than or equal to ((n+1)+1)/2=(n+2)/2?
    Offline

    11
    ReputationRep:
    (Original post by thomas575)
    Thanks, think this may get me there (i really hope so, sick of this lol). is it fair to say then that for the sum as above from 1 to 2^n then, when i do the inductive step i should get that it is greater than or equal to ((n+1)+1)/2=(n+2)/2?
    Yes, this is what you are trying to show (i.e. it is the end result).
 
 
 
Reply
Submit reply
Turn on thread page Beta
Updated: January 23, 2010

University open days

  1. University of Bradford
    University-wide Postgraduate
    Wed, 25 Jul '18
  2. University of Buckingham
    Psychology Taster Tutorial Undergraduate
    Wed, 25 Jul '18
  3. Bournemouth University
    Clearing Campus Visit Undergraduate
    Wed, 1 Aug '18
Poll
How are you feeling in the run-up to Results Day 2018?
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

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

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