Results 1 to 6 of 6

Math Help - More sequence help

  1. #1
    Junior Member
    Joined
    Mar 2009
    Posts
    31

    More sequence help

    Show that sequence r_n defined by the formula:

    r_n =  \frac {(-1)^n} {n!}

    satisfies the recurrence

    r_k =  \frac {-r_{k -1}}{k}


    This is another one where I really don't know where to start. Any help would be appreciated.

    THanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2009
    Posts
    44

    Smile my answer

    Quote Originally Posted by relyt View Post
    Show that sequence r_n defined by the formula:

    r_n =  \frac {(-1)^n} {n!}

    satisfies the recurrence

    r_k =  \frac {-r_{k -1}}{k}


    This is another one where I really don't know where to start. Any help would be appreciated.

    THanks
    This is a simple Qn .Isn`t it ?
    r_n =  \frac {(-1)^n} {n!}

    r_{n-1} =  \frac {(-1)^{n-1}} {{n-1}!}

    \frac {r_n}{r_{n-1}} = \frac { \frac {(-1)^n} {n!}}{\frac {(-1)^{n-1}} {(n-1)!}}

    then
    \frac {r_n}{r_{n-1}} = \frac{-1}{n}


    so it satisfies the above recurrence formula .
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Mar 2009
    Posts
    256
    Thanks
    1
    Quote Originally Posted by relyt View Post
    Show that sequence r_n defined by the formula:

    r_n = \frac {(-1)^n} {n!}

    satisfies the recurrence

    r_k = \frac {-r_{k -1}}{k}


    This is another one where I really don't know where to start. Any help would be appreciated.

    THanks
    .

    We have:

    r_{n-1} =\frac{\frac{(-1)^n}{(-1)}}{(n-1)!}=-\frac{(-1)^n}{(n-1)!}.................................................. ...........................1

    Now multiply (1) by 1/n and we get:

    -\frac{r_{n-1}}{n} = \frac{(-1)^n}{n(n-1)!} = r_{n}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Mar 2009
    Posts
    31
    Thanks to both of you for answering, but which is correct? I seem to get lost about half way though
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2009
    Posts
    44

    Thumbs up

    Quote Originally Posted by relyt View Post
    Thanks to both of you for answering, but which is correct? I seem to get lost about half way though
    both answers are correct bro ,
    but I think my answer is more descriptive .
    Have you not get it yet ?
    simply starts from r_n and r_{n-1} do the proof .
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Mar 2009
    Posts
    31
    Ok, i think I understand...I'll try some more

    THanks
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: August 24th 2010, 02:10 AM
  2. Replies: 0
    Last Post: July 4th 2010, 12:05 PM
  3. Replies: 2
    Last Post: March 1st 2010, 11:57 AM
  4. sequence membership and sequence builder operators
    Posted in the Discrete Math Forum
    Replies: 0
    Last Post: June 4th 2009, 03:16 AM
  5. Replies: 12
    Last Post: November 15th 2006, 12:51 PM

Search Tags


/mathhelpforum @mathhelpforum