Results 1 to 5 of 5

Thread: Divergent fractional series

  1. #1
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288

    Divergent fractional series

    Given that $\displaystyle \sum{a_n}$ diverges, and $\displaystyle a_n\geq{0}$, prove $\displaystyle \sum{\frac{a_n}{a_n+1}$ diverges

    So for any given $\displaystyle M\in{\mathbb{R}}, \exists N\in{\mathbb{N}}$ such that when $\displaystyle n\geq{N}$, $\displaystyle \sum^n_{k=1}{a_k}\geq{M}$

    Where do we go from here?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,742
    Thanks
    2814
    Awards
    1

    Re: Divergent fractional series

    Quote Originally Posted by I-Think View Post
    Given that $\displaystyle \sum{a_n}$ diverges, and $\displaystyle a_n\geq{0}$, prove $\displaystyle \sum{\frac{a_n}{a_n+1}$ diverges
    So for any given $\displaystyle M\in{\mathbb{R}}, \exists N\in{\mathbb{N}}$ such that when $\displaystyle n\geq{N}$, $\displaystyle \sum^n_{k=1}{a_k}\geq{M}$ Where do we go from here?
    You have two cases: $\displaystyle (a_n)\not\to 0$ and $\displaystyle (a_n)\to 0$.

    Consider the implications of each case.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor chisigma's Avatar
    Joined
    Mar 2009
    From
    near Piacenza (Italy)
    Posts
    2,162
    Thanks
    6

    Re: Divergent fractional series

    If $\displaystyle \lim_{n \rightarrow \infty} a_{n} \ne 0$ , then also $\displaystyle \lim_{n \rightarrow \infty} \frac{a_{n}}{1+a_{n}} \ne 0$ and the series $\displaystyle \sum_{n} \frac{a_{n}}{1+a_{n}}$ diverges. If $\displaystyle \lim_{n \rightarrow \infty} a_{n} = 0$ , then for n 'large enough' is $\displaystyle \frac{a_{n}}{1+a_{n}}> \frac{a_{n}}{2}$ and, because the series $\displaystyle \sum_{n} \frac{a_{n}}{2}$ diverges, also the series $\displaystyle \sum_{n} \frac{a_{n}}{1+a_{n}}$ diverges...

    Kind regards

    $\displaystyle \chi$ $\displaystyle \sigma$
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288

    Re: Divergent fractional series

    Okay, contemplated on hint, believe I got a solution

    For $\displaystyle a_n\rightarrow{x\neq{0}}$, $\displaystyle \frac{a_n}{1+a_n}\rightarrow{y\neq{0}}$, so result follows

    For $\displaystyle a_n\rightarrow{0}$, assume $\displaystyle \sum\frac{a_n}{1+a_n}$ convergent

    For any given $\displaystyle n$, let $\displaystyle Z=max(1+a_i)$, $\displaystyle 1\leq{i}\leq{n}$

    Choose $\displaystyle M$ so $\displaystyle \frac{M}{Z}>LUB(\sum\frac{a_n}{1+a_n})$, choose $\displaystyle N$ so that for $\displaystyle n\geq{N}$, $\displaystyle \sum{a_n}>M$

    Now $\displaystyle \sum{a_n}<Z\sum\frac{a_n}{1+a_n}<M$

    But $\displaystyle \sum{a_n}>M$, so a contradiction has occurred. The result follows

    Is this proof satisfactory?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member I-Think's Avatar
    Joined
    Apr 2009
    Posts
    288

    Re: Divergent fractional series

    Just saw chisigma's proof, it's slicker than mine, but I believe this is still correct, no?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. divergent series
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Jun 25th 2011, 10:42 PM
  2. Replies: 3
    Last Post: Feb 9th 2011, 04:35 AM
  3. Replies: 3
    Last Post: Mar 29th 2010, 11:36 AM
  4. Replies: 6
    Last Post: Feb 21st 2010, 04:52 AM
  5. why is this series divergent?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 23rd 2009, 07:51 PM

Search Tags


/mathhelpforum @mathhelpforum