Results 1 to 8 of 8

Math Help - Condesing Summations

  1. #1
    Member
    Joined
    Nov 2006
    From
    San Francisco
    Posts
    145

    Condesing Summations

    Okay, I have the following:

    <br />
\sum_{n=1}^{4}\frac{1}{n} + 1 + \sum_{n=1}^{3}\frac{1}{n} + 1 + \sum_{n=1}^{2}\frac{1}{n} + 1 + \sum_{n=1}^{1}\frac{1}{n}+1<br />

    I want to repeat this N times, so at any N I can find the total, so I suppose:

    <br />
\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N-1}\frac{1}{n}+....+\sum_{n=1}^{2}\frac{1}{n}+\sum_  {n=1}^{1}\frac{1}{n}+N<br />


    I want to write this in a single condensed sum (well N could be added on the end) or something that is at least simple, could someone help me? thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2005
    From
    Earth
    Posts
    1,599
    This is an interesting problem. From my thinking on it, I don't think it's possible to write this only using one sigma. The reason I think so is that you have too many things to keep track of and not enough variables in your current explanation. The total sum you seek is defined by other sums, so you must either define the smaller sums with another sigma or find a nice pattern in the sums that allows you to express the current total in terms of N.

    I thought of it like this:

    S = \left([N*\frac{1}{n}]+[(N-1)*\frac{1}{n+1}]...[(N-(N-1))*\frac{1}{N}] \right)+N and 1 < n < N.

    We must find a way to relate the increasing value of "n" and the decreasing value of "N" in terms of one variable. I can't think of a way to do it. Maybe some of these thoughts will be useful or someone else can solve it.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,392
    Thanks
    55
    Quote Originally Posted by jarny View Post
    Okay, I have the following:

    <br />
\sum_{n=1}^{4}\frac{1}{n} + 1 + \sum_{n=1}^{3}\frac{1}{n} + 1 + \sum_{n=1}^{2}\frac{1}{n} + 1 + \sum_{n=1}^{1}\frac{1}{n}+1<br />

    I want to repeat this N times, so at any N I can find the total, so I suppose:

    <br />
\sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N-1}\frac{1}{n}+....+\sum_{n=1}^{2}\frac{1}{n}+\sum_  {n=1}^{1}\frac{1}{n}+N<br />


    I want to write this in a single condensed sum (well N could be added on the end) or something that is at least simple, could someone help me? thanks
    I think the sum that you seek is (N+1)\sum_{n=1}^N \frac{1}{n}
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member redsoxfan325's Avatar
    Joined
    Feb 2009
    From
    Swampscott, MA
    Posts
    943
    Quote Originally Posted by Jameson View Post
    This is an interesting problem. From my thinking on it, I don't think it's possible to write this only using one sigma. The reason I think so is that you have too many things to keep track of and not enough variables in your current explanation. The total sum you seek is defined by other sums, so you must either define the smaller sums with another sigma or find a nice pattern in the sums that allows you to express the current total in terms of N.

    I thought of it like this:

    S = \left([N*\frac{1}{n}]+[(N-1)*\frac{1}{n+1}]...[(N-(N-1))*\frac{1}{N}] \right)+N and 1 < n < N.

    We must find a way to relate the increasing value of "n" and the decreasing value of "N" in terms of one variable. I can't think of a way to do it. Maybe some of these thoughts will be useful or someone else can solve it.
    Going off Jameson's idea, how about this:

    \sum_{n=1}^{N}\frac{N-n+1}{n}

    This is similar to Danny's, except that mine is \left((N+1)\sum_{n=1}^N\frac{1}{n}\right)-N

    Only one can be right; which is it?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Nov 2006
    From
    San Francisco
    Posts
    145
    Actually I was just pming Jameson, I got the same thing danny got actually hahaha I'm having trouble seeing why that is though. I was doing guess and check on my calculator and ran into the answer. I want to know why that is though :/
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,909
    Thanks
    770
    Hello, jarny!

    Okay, I have the following:

    \sum_{n=1}^{4}\frac{1}{n} + 1 + \sum_{n=1}^{3}\frac{1}{n} + 1 + \sum_{n=1}^{2}\frac{1}{n} + 1 + \sum_{n=1}^{1}\frac{1}{n}+1<br />

    I want to repeat this N times, so at any N, I can find the total, so I suppose:

    \sum_{n=1}^{N}\frac{1}{n}+\sum_{n=1}^{N-1}\frac{1}{n}+ \hdots +\sum_{n=1}^{2}\frac{1}{n}+\sum_{n=1}^{1}\frac{1}{  n}+N<br />

    I want to write this in a single condensed sum (N could be added on the end) .
    I wrote out the general list . . . and added.


    \sum^1_{k=1}\frac{1}{k} \;=\;\frac{1}{1}

    \sum^2_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2}

    \sum^3_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2} + \frac{1}{3}

    \sum^4_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}

    . . \vdots. . . . . . . . \vdots

    \sum^{n-1}_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots + \frac{1}{n-1}

    \sum^n_{k=1}\frac{1}{k} \;=\;\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \hdots + \frac{1}{n-1} + \frac{1}{n}



    \text{Sum} \;=\;n\left(\frac{1}{1}\right) + (n-1)\left(\frac{1}{2}\right) + (n-2)\left(\frac{1}{3}\right) + \hdots + 2\left(\frac{1}{n-1}\right) + 1\left(\frac{1}{n}\right)

    . . . = \;\frac{n}{1} + \frac{n-1}{2} + \frac{n-2}{3} + \frac{n-3}{4} + \hdots + \frac{2}{n-1} + \frac{1}{n}

    . . . = \;\sum^n_{k=1}\frac{n-k+1}{k}

    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Nov 2006
    From
    San Francisco
    Posts
    145
    Soroban, Your sum comes up with a different result...I'm trying to figure out what is wrong though :/
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Nov 2006
    From
    San Francisco
    Posts
    145
    Okay, Your way is much much easier, I think you just forgot to add n at the end. Thank you very much!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Summations
    Posted in the Discrete Math Forum
    Replies: 7
    Last Post: December 6th 2010, 08:22 AM
  2. summations
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 27th 2010, 05:23 AM
  3. Summations please help!!!
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: November 15th 2009, 02:01 AM
  4. summations
    Posted in the Algebra Forum
    Replies: 6
    Last Post: December 16th 2008, 02:41 PM
  5. Summations
    Posted in the Calculus Forum
    Replies: 0
    Last Post: October 26th 2008, 12:37 PM

Search Tags


/mathhelpforum @mathhelpforum