Results 1 to 5 of 5

Math Help - Math Help Right Away!!!

  1. #1
    Junior Member
    Joined
    Jan 2006
    From
    Oakland
    Posts
    55

    Exclamation Math Help Right Away!!!

    2. (n!)/(n-2!) n= 2,3,4... Converges or Diverges? If converges where does it converge?

    3. Summation of n=0 goes to infinity of the equation 2(-1/2)^n Converges where?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by Nimmy
    2. (n!)/(n-2!) n= 2,3,4... Converges or Diverges? If converges where does it converge?

    3. Summation of n=0 goes to infinity of the equation 2(-1/2)^n Converges where?
    Hello,

    to 2.) I hope you meant \frac{n!}{(n-2)!}. Then you can transform this term to:
    \frac{2\cdot3\cdot4\cdot...\cdot (n-2) \cdot (n-1)\cdot n}{2 \cdot3 \cdot 4 \cdot...\cdot (n-2)}
    and thats (n-1)*n which doesn't converge for n to infinity.

    to 3. Split the summation into two sums: In 1st sum n is even and the numbers are positive, in the 2nd sum n is odd and all numbers are negative. Calculate both sums separately.

    This sum must converge according to the convergence criterium of Leibniz.

    Greetings

    EB
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    The series,
    \sum^{\infty}_{k=0}2(-1/2)^k.
    Is the same as 2\sum^{\infty}_{k=0}(-1/2)^k. Further, it is a geometric series with constant ratio -1/2. Thus, not only it converges we also can easily determine to what it converges to, which is 3/2.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Jan 2006
    From
    Oakland
    Posts
    55

    Exclamation

    Quote Originally Posted by ThePerfectHacker
    The series,
    \sum^{\infty}_{k=0}2(-1/2)^k.
    Is the same as 2\sum^{\infty}_{k=0}(-1/2)^k. Further, it is a geometric series with constant ratio -1/2. Thus, not only it converges we also can easily determine to what it converges to, which is 3/2.
    Ok Let me rephrase the question just to make sure the ppl in the fourm understand

    Find the sum of the geometric series:

    \sum^{\infty}_{k=0}2(-1/2)^k= 2-1+1/2-1/4...

    can u explain how you got it to converge at 3/2?
    Last edited by Nimmy; February 23rd 2006 at 03:36 PM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    9
    Quote Originally Posted by Nimmy
    Ok Let me rephrase the question just to make sure the ppl in the fourm understand

    Find the sum of the geometric series:

    \sum^{\infty}_{k=0}2(-1/2)^k= 2-1+1/2-1/4...

    can u explain how you got it to converge at 3/2?
    Given the series,
    \sum^{\infty}_{k=0}r^k
    It converges if and only if, |r|<1.
    Now if |r|<1 then it converges to,
    \frac{1}{1-r}.
    That is the property of the infinite geometric series.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 0
    Last Post: May 29th 2011, 12:57 AM
  2. Replies: 2
    Last Post: February 14th 2011, 02:31 AM
  3. Replies: 1
    Last Post: June 19th 2010, 10:05 PM
  4. Replies: 1
    Last Post: June 2nd 2010, 09:49 AM
  5. Replies: 2
    Last Post: February 15th 2010, 03:09 PM

Search Tags


/mathhelpforum @mathhelpforum