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Establishing an inequality

  1. Nov 8, 2005 #1
    I'm working on a series but since the thing that I need help with is only a simple inequality, this seems like the appropriate subsection to post this thread in.

    \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^n \frac{{\log \left( n \right)}}{{\sqrt n }}}

    With the limits I've pretty much been given, the limit of the terms is zero and the terms are greater than zero for all finite n. I've got a feeling that the series does converge. So I need to show that a_n >= a_(n+1) for all n, or that the inequality holds from a certain point anyway.

    \frac{{\log \left( n \right)}}{{\sqrt n }} \ge \frac{{\log \left( {n + 1} \right)}}{{\sqrt {n + 1} }}

    I can't think of a way to explicitly show the above inequality. I thought about rewriting the argument of the logarithm as square root but that doesn't appear to lead anywhere.

    Any help would be great thanks.
  2. jcsd
  3. Nov 9, 2005 #2
    Since I only need to show that the inequality holds after a certain value of n, I think taking the limit as n goes to infinity might help. But to make use of that I think I need to split the original expression in a certain way. Does anyone have any suggestions?

    Edit: I can think of a way to get to the answer using an indirect approach. But going that way is a clumsy approach. I am wondering if there is a more direct way(taking limits to say some expression is less than 1 for sufficiently large n is ok) to establish the inequality.
    Last edited: Nov 9, 2005
  4. Nov 9, 2005 #3


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    You just said the limit of the terms is 0 and this is an alternating series! That tells you everything you need to know!
  5. Nov 9, 2005 #4
    The thing is that from what I've been told, an alternating series converges only if the following three conditions are satisfied.

    (i) Limit of terms is zero.
    (ii) The terms are greater than zero for all finite n.
    (iii) a_n >= a_(n+1) for all n, or after a certain number of finite values.

    The limit of the terms is zero from a standard limit I can quote. The a_n terms(ie the expression in the series without the alternating part) are clearly greater than zero for all finite n. However, what is not so clear (algebraically, not intuitively) that the terms are decreasing.

    To show that the terms(excluding (-1)^n which is the alternating part) are decreasing I need to show that the inequality with the log holds true for all n, or all n sufficiently large. I'm finding it to be rather difficult to justify the inequality (the one in my first post) without a little bit of fudging with the limits. I just really wanted to see if there was a more concise way of showing that the terms are decreasing using a bit of algebra. As an example what I mean by showing that the terms of an alternating series are decreasing here is one.

    \sum\limits_{n = 1}^\infty {\left( { - 1} \right)^n \frac{{1 + 3^n }}{{1 + 4^n }}}

    a_n = \frac{{1 + 3^n }}{{1 + 4^n }}

    Clearly, a_n > 0 for all finite n and the limit of the terms is zero. So we need to show algebraically that a_n >= a_(n+1) for all finite n.

    a_n = \frac{{\frac{1}{{4^n }} + \left( {\frac{3}{4}} \right)^n }}{{\frac{1}{{4^n }} + 1}}

    a_{n + 1} = \frac{{1 + 3^{n + 1} }}{{1 + 4^{n + 1} }} = \frac{{\frac{1}{{4^{n + 1} }} + \left( {\frac{3}{4}} \right)^{n + 1} }}{{\left( {\frac{1}{4}} \right)^{n + 1} + 1}}

    I'd make a few comments as well but the point I'm trying to make is that this explicitly shows that the terms are decreasing which is the kind of thing I needed to do with the log series.
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