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Thread: 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 $\displaystyle 0<a_{n+1}<a_{n}$ for all $\displaystyle n$.

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

    hence show that if this series

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

    hence determine if these series converge or diverge
    $\displaystyle \sum \frac {1}{n \log n}$ and $\displaystyle \sum \frac {1} {n(\log n)^2}$
    Last edited by iLikeMaths; Jan 22nd 2009 at 11:27 PM.
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  2. #2
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    Convergence

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

    $\displaystyle \frac {1}{2}(2^{n+1} a_{2^n+1})<a_{2^n+1}+...<2^n a_{2^n}$ where the expression in the middle (the one with the dots ...) actually means $\displaystyle a_{2^n+1}+... a_{2^{n+1}}$ (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: $\displaystyle (a_9 + ... + a_{16})$

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

    $\displaystyle a_{2^3+1} + ... + a_{2^4}$

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

    Generalise this to $\displaystyle n$, and you get

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

    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 $\displaystyle \sum a_n$, I think you then use the fact that the sequence $\displaystyle a_{2^n+1}+... a_{2^{n+1}}$ 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 $\displaystyle 0<a_{n+1}<a_{n}$ for all $\displaystyle n$.

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

    hence show that if this series

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

    hence determine if these series converge or diverge
    $\displaystyle \sum \frac {1}{n \log n}$ and $\displaystyle \sum \frac {1} {n(\log n)^2}$
    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 $\displaystyle 0 < a_{n+1} < a_n$.

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

    Since $\displaystyle \frac{a_{n+1}}{a_n} < 1$ 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 $\displaystyle 0 < a_{n+1} < a_n$.

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

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


    Can anyone see any flaws to my logic?
    Just a quick reply to this post. I think it has to be $\displaystyle \lim_{n \rightarrow \infty}| \frac{a_{n+1}}{a_n}| < 1$ 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 $\displaystyle \sum_{i=0}^\infty a_i$ as follows:

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

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

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

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

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

    which is the RHS of the first inequality.

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

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

    So we have now shown that:

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

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

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

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

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

    Grandad
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