Results 1 to 5 of 5

Thread: Series - nth term - Is this right?

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    4

    Series - nth term - Is this right?

    I am not 100% confident about this so would someone mind checking my work?

    The nth partial sum of a Series is given by:
    $\displaystyle Sn=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\ frac{1}{2n}=\sum_{k=1}^{n}\frac{1}{n+k}$

    1. What is the associated Series:

    $\displaystyle \sum_{n=1}^{\infty}\frac{1}{2n}$

    2. Convergence?

    $\displaystyle \lim_{n\to\infty}\frac{1}{2n}=0$

    General term goes to 0

    Test for convergence by direct comparison.
    $\displaystyle \frac{1}{2n}=\frac{1}{n-\frac{1}{2}}>\frac{1}{n}$

    $\displaystyle \frac{1}{n}$ divergent harmonic series

    $\displaystyle \sum_{n=1}^{\infty}\frac{1}{2n}$ diverges by direct comparison with $\displaystyle \frac{1}{n}$

    3. Determine the Limit if possible

    $\displaystyle \lim_{n\to\infty}\;Sn=\lim_{n\to\infty}\frac{1}{2n }=0$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by tiredNConfused View Post
    I am not 100% confident about this so would someone mind checking my work?

    The nth partial sum of a Series is given by:
    $\displaystyle Sn=\frac{1}{n+1}+\frac{1}{n+2}+\frac{1}{n+3}+...+\ frac{1}{2n}=\sum_{k=1}^{n}\frac{1}{n+k}$

    1. What is the associated Series:
    Let $\displaystyle a_n = \left\{ \begin{array}{c} \frac{1}{2} \text{ for }n=1 \\ \frac{1}{2n-1}+\frac{1}{2n} \text{ for }n\geq 2 \end{array} \right.$

    Then the series is $\displaystyle \sum_{n=1}^{\infty} a_n$

    This series converges if and only if $\displaystyle \sum_{n=2}^{\infty} a_n = \sum_{n=2}^{\infty} \left( \frac{1}{2n-1} + \frac{1}{2n} \right)$ converges.

    Can you finish this now?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Oct 2008
    Posts
    4
    Yes I can, I knew it wasn't as easy as I made it.

    But one question, how did you come up with: $\displaystyle \frac{1}{2n-1}$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by tiredNConfused View Post
    Yes I can, I knew it wasn't as easy as I made it.

    But one question, how did you come up with: $\displaystyle \frac{1}{2n-1}$
    Because,
    $\displaystyle s_1 = \boxed{a_1 = \tfrac{1}{2}}$
    $\displaystyle s_2 = \boxed{a_1+a_2 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}}$
    $\displaystyle s_3 = \boxed{a_1+a_2+a_3 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}+\tfrac{1}{5 }+\tfrac{1}{6}}$

    Now solve the first boxed equation, well it is already solved, $\displaystyle a_1 = \tfrac{1}{2}$.

    For the second boxed equation we have $\displaystyle a_1 + a_2 = \tfrac{1}{2}+\tfrac{1}{3}+\tfrac{1}{4}$ and we know what $\displaystyle a_1$ is.
    Solving it gives, $\displaystyle a_2 = \tfrac{1}{3}+\tfrac{1}{4}$.

    Similar with the third boxed equation we get $\displaystyle a_3 = \tfrac{1}{5} + \tfrac{1}{6}$

    And this generalizes. That is how we get the formula for $\displaystyle a_1$.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    4
    Thanks soooo much!!!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Show that p/q is the ...th term of the series...
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: Feb 12th 2011, 09:10 PM
  2. Power series, term-by-term integration
    Posted in the Calculus Forum
    Replies: 3
    Last Post: Apr 8th 2010, 02:14 AM
  3. term of a series
    Posted in the Algebra Forum
    Replies: 1
    Last Post: Feb 26th 2009, 01:14 AM
  4. find kth term in a series
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Feb 22nd 2009, 03:59 PM
  5. nth term for series
    Posted in the Algebra Forum
    Replies: 3
    Last Post: Apr 8th 2007, 05:35 AM

Search Tags


/mathhelpforum @mathhelpforum