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Math Help - convergence

  1. #1
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    convergence

    I have been having trouble with this question for days and i am having difficulty understanding what the question is asking me, its on the subject of series convergence and divergence, its a long question

    This question concerns grouping the terms of a series as follows:
    a0+(a1+a2)+(a3+a4)+(a5+ +a8)+(a9+ +a16)+(a17+ +a32)+

    Suppose that  0<a_{n+1}<a_{n} for all n.

    Show that
    \frac {1}{2}(2^{n+1} a_{2n+1})<a_{2n+1}+....<2^n a_{2n}

    hence show that if this series

    \sum a_n converges then this series \sum 2^n a_2n also converges. and that if one of the two series diverges then so does the other.

    hence determine if these series converge or diverge
    \sum \frac {1}{n \log n} and \sum \frac {1} {n(\log n)^2}
    Last edited by iLikeMaths; January 23rd 2009 at 12:27 AM.
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  2. #2
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    Convergence

    Hello iLikeMaths
    Quote Originally Posted by iLikeMaths View Post
    Show that
    \frac {1}{2}(2^{n+1} a_{2n+1})<a_{2n+1}+....<2^n a_{2n}
    Are you sure this shouldn't be:

    \frac {1}{2}(2^{n+1} a_{2^n+1})<a_{2^n+1}+...<2^n a_{2^n} where the expression in the middle (the one with the dots ...) actually means a_{2^n+1}+... a_{2^{n+1}} (and not the sum to infinity which I originally took it to mean).

    The reason I ask is that this ties in with the preamble to the question. For instance, look at the terms in the fourth set of brackets: (a_9 + ... + a_{16})

    This corresponds to n = 3, 2^n = 8, 2^n + 1 = 9. So it is the sequence

    a_{2^3+1} + ... + a_{2^4}

    There are 2^3 = 8 terms in this bracket, each of which is less than the last term in the previous bracket, a_8 (= a_{2^3}) because a_{n+1}<a_n for all n, and so their sum is less than 2^3 \times a_{2^3}

    Generalise this to n, and you get

    \sum_{i=2^n+1}^{2^{n+1}}a_i < 2^na_{2^n}

    which I think is what you want for the right-hand side of the inequality. The left-hand side is similar, but not quite the same.

    To look at the convergence of \sum a_n, I think you then use the fact that the sequence a_{2^n+1}+... a_{2^{n+1}} is squeezed in between the other two.

    Does that give you some ideas?

    Grandad
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  3. #3
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    yes your post has been helpful, and i forgot to put the expression in the middle, could you please explain in more detail, because i am totally lost on the whole question, thank you
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  4. #4
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    Quote Originally Posted by iLikeMaths View Post
    I have been having trouble with this question for days and i am having difficulty understanding what the question is asking me, its on the subject of series convergence and divergence, its a long question

    This question concerns grouping the terms of a series as follows:
    a0+(a1+a2)+(a3+a4)+(a5+ +a8)+(a9+ +a16)+(a17+ +a32)+

    Suppose that  0<a_{n+1}<a_{n} for all n.

    Show that
    \frac {1}{2}(2^{n+1} a_{2n+1})<a_{2n+1}+....<2^n a_{2n}

    hence show that if this series

    \sum a_n converges then this series \sum 2^n a_2n also converges. and that if one of the two series diverges then so does the other.

    hence determine if these series converge or diverge
    \sum \frac {1}{n \log n} and \sum \frac {1} {n(\log n)^2}
    I haven't studied sequences and series in quite a while, but couldn't you just use a ratio test to determine its convergence or divergence?

    You're told 0 < a_{n+1} < a_n.

    So a_{n + 1} < a_n and thus \frac{a_{n+1}}{a_n} < 1 (we can divide and know it's not going to change the inequality because we're told both terms are positive).

    Since \frac{a_{n+1}}{a_n} < 1 the series converges.


    Can anyone see any flaws to my logic?
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  5. #5
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    Ratio test

    Hello Prove It
    Quote Originally Posted by Prove It View Post
    I haven't studied sequences and series in quite a while, but couldn't you just use a ratio test to determine its convergence or divergence?

    You're told 0 < a_{n+1} < a_n.

    So a_{n + 1} < a_n and thus \frac{a_{n+1}}{a_n} < 1 (we can divide and know it's not going to change the inequality because we're told both terms are positive).

    Since \frac{a_{n+1}}{a_n} < 1 the series converges.


    Can anyone see any flaws to my logic?
    Just a quick reply to this post. I think it has to be \lim_{n \rightarrow \infty}| \frac{a_{n+1}}{a_n}| < 1 for a convergent sequence, and we don't know this.

    Haven't time to write more at present. Can anyone else take this up?

    Grandad
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  6. #6
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    Convergence

    Hello iLikeMaths

    Here's a little more help.

    To understand how this works, you must study carefully the way in which the brackets that group together the terms are formed. So, we group the sequence \sum_{i=0}^\infty a_i as follows:

    a_0 + a_1 + a_2 + (a_3 + a_4) + (a_5 + ... + a_8) + (a_9 + ... + a_{16}) + (a_{17}+ ... +a_{32}) + ...

    In this grouping, then, after the first three terms a_0, a_1 and a_2, the number of terms in each bracket is 2, 4, 8, 16, ..., the first term in each bracket is a_3, a_5, a_9, a_{17}, ... and the last term in each bracket is a_4, a_8, a_{16}, a_{32}, ...

    Thus the n^{th} bracket contains 2^n terms, the first term being a_{2^n+1} and last term being a_{2^{n+1}}

    Now this is a decreasing sequence of positive numbers; that is to say 0<a_{n+1}<a_n for all n. So all the terms in the n^{th} bracket are smaller than the last term in the (n-1)^{th} bracket. And there are 2^n terms in the n^{th} bracket. So:

    a_{2^n+1} + a_{2^n+2} + ... + a_{2^{n+1}} < a_{2^n} + a_{2^n} + ... + a_{2^n} = 2^na_{2^n}

    which is the RHS of the first inequality.

    To get the LHS of this inequality, note that each term except the last in the n^{th} bracket is greater than the last term in this bracket, which is a_{2^{n+1}}. So:

    a_{2^n+1} + a_{2^n+2} + ... + a_{2^{n+1}} > a_{2^{n+1}} + a_{2^{n+1}} + ... + a_{2^{n+1}} = 2^na_{2^{n+1}}= \frac{1}{2}(2^{n+1}a_{2^{n+1}})

    So we have now shown that:

    \frac{1}{2}(2^{n+1}a_{2^{n+1}}) < a_{2^n+1} + a_{2^n+2} + ... + a_{2^{n+1}}<2^na_{2^n}

    So when we sum all the parts of the inequality, we get:

    \frac{1}{2}\sum_{n=1}^\infty(2^{n+1}a_{2^{n+1}}) < \sum_{n=3}^\infty a_n < \sum_{n=1}^\infty 2^na_{2^n}

    \Rightarrow \frac{1}{2}\sum_{n=2}^\infty(2^{n}a_{2^{n}}) < \sum_{n=3}^\infty a_n < \sum_{n=1}^\infty 2^na_{2^n}

    Can you understand all this? Can you see how this leads to the statements about convergence and divergence?

    Grandad
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