Results 1 to 6 of 6

Math Help - Summations please help!!!

  1. #1
    Senior Member
    Joined
    Apr 2009
    Posts
    308

    Summations please help!!!

    1. Find a formula for the sum

    2. Find the sum of

    3. Find a formula for

    Can anyone please solve step by step, I just started summations and need to see the procedure of doing these questions

    Thanks!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Eater of Worlds
    galactus's Avatar
    Joined
    Jul 2006
    From
    Chaneysville, PA
    Posts
    3,001
    Thanks
    1
    For #1: It can be rewritten as \sum_{k=1}^{\infty}\frac{1}{k!}-\sum_{k=1}^{\infty}\frac{1}{(k+1)!}

    The first one, the sum of the reciprocals of the factorials, is a well-known sum which equals e-1 and can be derived from the Taylor series for e by letting x=1.

    The second one \sum_{k=1}^{\infty}\frac{1}{(k+1)!}=e-2

    e-1-(e-2)=1
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by usagi_killer View Post
    1. Find a formula for the sum

    2. Find the sum of


    \log_ku=\frac{\ln u}{\ln k}\Longrightarrow \sum\limits_{k=2}^n\frac{1}{\log_ku}= \frac{1}{\ln u}\sum\limits_{k=2}^n\ln k=\frac{1}{\ln u}\ln\left(\prod\limits_{k=2}^nk\right)=\frac{\ln k!}{\ln u}=\log_uk!

    Tonio


    3. Find a formula for

    Can anyone please solve step by step, I just started summations and need to see the procedure of doing these questions

    Thanks!
    .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    13
    Quote Originally Posted by usagi_killer View Post
    3. Find a formula for
    \sum\limits_{j=1}^{n}{\frac{1}{j(j+1)(j+2)}}=\frac  {1}{2}\sum\limits_{j=1}^{n}{\frac{(j+2)-j}{j(j+1)(j+2)}}, hence your sum equals \frac{1}{2}\sum\limits_{j=1}^{n}{\left( \frac{1}{j(j+1)}-\frac{1}{(j+1)(j+2)} \right)}=\frac{1}{2}\left( \frac{1}{2}-\frac{1}{(n+1)(n+2)} \right).
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member
    Joined
    Apr 2009
    Posts
    308
    Quote Originally Posted by galactus View Post
    For #1: It can be rewritten as \sum_{k=1}^{\infty}\frac{1}{k!}-\sum_{k=1}^{\infty}\frac{1}{(k+1)!}

    The first one, the sum of the reciprocals of the factorials, is a well-known sum which equals e-1 and can be derived from the Taylor series for e by letting x=1.

    The second one \sum_{k=1}^{\infty}\frac{1}{(k+1)!}=e-2

    e-1-(e-2)=1
    Thanks for the help! Is there a way to do this without Taylor's series and how did you get the 2nd summation?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Banned
    Joined
    Oct 2009
    Posts
    4,261
    Thanks
    2
    Quote Originally Posted by usagi_killer View Post
    Thanks for the help! Is there a way to do this without Taylor's series and how did you get the 2nd summation?

    Galactus did a nice job...but with an infinite series, whereas you wanted a finite one, yet the answer STILL is in Galactus message:

    \sum\limits_{k=1}^n\frac{k}{(k+1)!}=\sum\limits_{k  =1}^n\frac{1}{k!}-\sum\limits_{k=1}^n\frac{1}{(k+1)!}=1-\frac{1}{(n+1)!}=\frac{(n+1)!-1}{(n+1)!}

    Tonio
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Help with summations.
    Posted in the Algebra Forum
    Replies: 1
    Last Post: January 25th 2011, 11:37 AM
  2. Summations
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: December 6th 2010, 07:22 AM
  3. summations
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 27th 2010, 04:23 AM
  4. summations
    Posted in the Algebra Forum
    Replies: 6
    Last Post: December 16th 2008, 01:41 PM
  5. Summations
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 26th 2008, 11:37 AM

Search Tags


/mathhelpforum @mathhelpforum