Results 1 to 7 of 7

Math Help - mathematical induction problem

  1. #1
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    203

    mathematical induction problem

    Hi

    I am trying to prove the following by mathematical induction.

    2\sum_{k=1}^n \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{n^2} \right)

    \forall \; n \;\;\geqslant  2

    first step is to test this for n=2, which I did. If we call above statement P(n) , then
    P(2) is true. So we assume that let the statement be true for some
    m\;\in N with m>2 and the statement is also true for
    i=2,3, \cdots ,m . So we have

    2\sum_{k=1}^m \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{m^2} \right)\cdots (1)

    and such statements from i=2 to i=m. We have to prove that the statement is true for P(m+1). i.e.

    2\sum_{k=1}^{m+1} \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{(m+1)^2} \right)

    So consider

    2\sum_{k=1}^{m+1} \;\frac{1}{k^3}

    =2\sum_{k=1}^m \;\frac{1}{k^3}+\frac{2}{(m+1)^3}

    Now we can use equation 1 here

    < \;\left( 3-\frac{1}{m^2}\right) +\frac{2}{(m+1)^3}

    I tried to manipulate the second term here. We see that for m > 2 , we have

    (m+1)^3 > 2(m+1)^2}

    \Rightarrow \;  \frac{1}{(m+1)^3} < \frac{1}{2(m+1)^2}

    \Rightarrow \; \frac{2}{(m+1)^3} < \frac{1}{(m+1)^2}

    so using this above, we have

    2\sum_{k=1}^{m+1} \;\frac{1}{k^3} < \left( 3-\frac{1}{m^2} \right)+\frac{1}{(m+1)^2}

    I don't know where to go from here....... can anybody suggest something ?

    thanks
    Last edited by issacnewton; June 6th 2011 at 11:27 PM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    203
    The last equation should be

    2\sum_{k=1}^{m+1} \;\frac{1}{k^3} < \left( 3-\frac{1}{m^2} \right)+\frac{1}{(m+1)^2}

    for some reasons the Edit button didn't work properly. It couldn't save the changed matter. So I am replying to my own post.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    I would try

    P(m)

    2\sum_{k=1}^m\frac{1}{k^3}<3-\frac{1}{m^2}


    P(m+1)

    2\sum_{k=1}^{m+1}\frac{1}{k^3}<3-\frac{1}{(m+1)^2}


    Then if P(m) is true

    2\sum_{k=1}^{m+1}\frac{1}{k^3}<3-\frac{1}{m^2}+\frac{2}{(m+1)^3}

    Then P(m+1) will DEFINATELY be true if

    3-\frac{1}{(m+1)^2}>3-\frac{1}{m^2}+\frac{2}{(m+1)^3}

    Then to prove...

    \frac{2}{(m+1)^3}+\frac{1}{(m+1)^2}-\frac{1}{m^2}<0\;\;\;?

    \frac{2+m+1}{(m+1)^3}<\frac{1}{m^2}\;\;\;?

    \frac{3m^2+m^3}{(m+1)^3}<1\;\;\;?

    which is true if you expand the denominator.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,408
    Thanks
    1294
    Quote Originally Posted by issacnewton View Post
    The last equation should be

    2\sum_{k=1}^{m+1} \;\frac{1}{k^3} < \left( 3-\frac{1}{m^2} \right)+\frac{1}{(m+1)^2}

    for some reasons the Edit button didn't work properly. It couldn't save the changed matter. So I am replying to my own post.
    For some reason editing only works if you "Go Advanced".
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    203
    Archie, thanks .... but this is not the usual way its done.. I spent lot of time thinking about this....
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Joined
    Dec 2009
    Posts
    3,120
    Thanks
    1
    Quote Originally Posted by issacnewton View Post
    Hi

    I am trying to prove the following by mathematical induction.

    2\sum_{k=1}^n \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{n^2} \right)

    \forall \; n \;\;\geqslant  2

    first step is to test this for n=2, which I did. If we call above statement P(n) , then
    P(2) is true. So we assume that let the statement be true for some
    m\;\in N with m>2 and the statement is also true for
    i=2,3, \cdots ,m . So we have

    2\sum_{k=1}^m \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{m^2} \right)\cdots (1)

    and such statements from i=2 to i=m. We have to prove that the statement is true for P(m+1). i.e.

    2\sum_{k=1}^{m+1} \;\; \frac{1}{k^3}< \;\; \left( 3-\frac{1}{(m+1)^2} \right)

    So consider

    2\sum_{k=1}^{m+1} \;\frac{1}{k^3}

    =2\sum_{k=1}^m \;\frac{1}{k^3}+\frac{2}{(m+1)^3}

    Now we can use equation 1 here

    < \;\left( 3-\frac{1}{m^2}\right) +\frac{2}{(m+1)^3}

    I tried to manipulate the second term here. We see that for m > 2 , we have

    (m+1)^3 > 2(m+1)^2}

    \Rightarrow \;  \frac{1}{(m+1)^3} < \frac{1}{2(m+1)^2}

    \Rightarrow \; \frac{2}{(m+1)^3} < \frac{1}{(m+1)^2}

    so using this above, we have

    2\sum_{k=1}^{m+1} \;\frac{1}{k^3} < \left( 3-\frac{1}{m^2} \right)+\frac{1}{(m+1)^2}

    I don't know where to go from here....... can anybody suggest something ?

    thanks
    From here, can you see you want to show that P(m+1) is absolutely true if

    3-\frac{1}{m^2}+\frac{1}{(m+1)^2}>3-\frac{1}{(m+1)^2}\;\;\;?

    Then

    \frac{2}{(m+1)^2}>\frac{1}{m^2}

    which can be written as

    m^2>2m+1\;\;?

    which is true for m=3 and above.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Oct 2010
    From
    Mumbai, India
    Posts
    203
    I did understand your original reasoning too. But this one is nicer too. Many problems in analysis basically boil down to playing with inequalities I guess
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Mathematical Induction problem
    Posted in the Discrete Math Forum
    Replies: 3
    Last Post: April 6th 2011, 11:52 AM
  2. Mathematical Induction Problem.
    Posted in the Differential Geometry Forum
    Replies: 14
    Last Post: June 9th 2010, 11:22 AM
  3. Mathematical Induction Problem!
    Posted in the Algebra Forum
    Replies: 2
    Last Post: April 15th 2010, 05:06 PM
  4. Mathematical Induction Problem
    Posted in the Calculus Forum
    Replies: 2
    Last Post: September 15th 2009, 05:38 PM
  5. mathematical induction problem
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: January 12th 2008, 07:22 PM

/mathhelpforum @mathhelpforum