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Math Help - does the sequence necessarily become decreasing??

  1. #1
    Senior Member abhishekkgp's Avatar
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    does the sequence necessarily become decreasing??

    Consider a sequence (x_n) of real numbers with the following properties:
    1) x_n>0 \, \forall \, n
    2) x_n<1 \, \forall \, n
    3) x_{n+1}>x_n \, \forall \, n

    Define the sequence (y_n) as y_n=\frac{x_n}{n}.

    Does (y_n) necessarily become decreasing after a sufficiently large value of n.

    In other words, Does there exist a N such that y_{n+1} \leq y_n \,  \forall \, n \, \geq N.

    This is not a textbook question so i don't know the answer myself.
    I can't think of an example of (x_n) such that y_{n+1} > y_n occurs frequently.
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  2. #2
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    Re: does the sequence necessarily become decreasing??

    Quote Originally Posted by abhishekkgp View Post
    Consider a sequence (x_n) of real numbers with the following properties:
    1) x_n>0 \, \forall \, n
    2) x_n<1 \, \forall \, n
    3) x_{n+1}>x_n \, \forall \, n

    Define the sequence (y_n) as y_n=\frac{x_n}{n}.

    Does (y_n) necessarily become decreasing after a sufficiently large value of n.

    In other words, Does there exist a N such that y_{n+1} \leq y_n \,  \forall \, n \, \geq N.

    This is not a textbook question so i don't know the answer myself.
    I can't think of an example of (x_n) such that y_{n+1} > y_n occurs frequently.
    The answer is No. In other words, such a sequence (y_n) does not necessarily eventually decrease.

    To start with, consider the sequence (x_n) defined by x_n = \tfrac k{k+1} for k^2+1\leqslant n\leqslant (k+1)^2\ (k=0,1,2,\ldots). Then

    y_{k^2+1} = \frac{x_{k^2+1}}{k^2+1} = \frac k{(k+1)(k^2+1)},\qquad y_{k^2} = \frac{x_{k^2}}{k^2} = \frac {k-1}{k^3},

    and you can check by multiplying out the fractions that y_{k^2+1} > y_{k^2} for all k. Thus there are infinitely many values of n for which y_{n+1}>y_n.

    That example does not quite answer the original question, because the sequence (x_n) is not strictly increasing (being constant throughout the interval k^2+1\leqslant n\leqslant (k+1)^2). However, in principle there is no difficulty in adjusting the sequence so as to make it strictly increasing. For each k, you can make x_{k^2+1} very slightly smaller, without disturbing the property that y_{k^2+1} > y_{k^2}. Then you can interpolate the values of x_n linearly for k^2+1\leqslant n\leqslant (k+1)^2 so as to ensure that the sequence increases strictly.
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  3. #3
    MHF Contributor chisigma's Avatar
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    Re: does the sequence necessarily become decreasing??

    Quote Originally Posted by abhishekkgp View Post
    Consider a sequence (x_n) of real numbers with the following properties:
    1) x_n>0 \, \forall \, n
    2) x_n<1 \, \forall \, n
    3) x_{n+1}>x_n \, \forall \, n

    Define the sequence (y_n) as y_n=\frac{x_n}{n}.

    Does (y_n) necessarily become decreasing after a sufficiently large value of n.

    In other words, Does there exist a N such that y_{n+1} \leq y_n \, \forall \, n \, \geq N.

    This is not a textbook question so i don't know the answer myself.
    I can't think of an example of (x_n) such that y_{n+1} > y_n occurs frequently.
    If we define \delta_{x}(n)= x_{n+1}-x_{n} then it must be...

    a) \delta_{x}(n)>0\ \forall n

    b) \lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1 (1)

    Now, if y_{n}=\frac{x_{n}}{n} we define \delta_{y}(n)= y_{n+1}-y_{n} and in few steps we find that...

    \delta_{y}(n)= \frac{\delta_{x}(n)-y_{n}}{n+1} (2)

    Because the series in (1) converges, it must be necesserly for some \varepsilon>0 and N \delta_{x}(n)<\frac{\varepsilon}{n}\ \forall n>N and that means that \delta_{x}(n)<y_{n}\ \forall n>N so that \forall n>N is \delta_{y}(n)<0...

    Kind regards

    \chi \sigma
    Last edited by chisigma; December 28th 2011 at 11:52 PM.
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  4. #4
    Senior Member abhishekkgp's Avatar
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    Re: does the sequence necessarily become decreasing??

    Quote Originally Posted by Opalg View Post
    The answer is No. In other words, such a sequence (y_n) does not necessarily eventually decrease.

    To start with, consider the sequence (x_n) defined by x_n = \tfrac k{k+1} for k^2+1\leqslant n\leqslant (k+1)^2\ (k=0,1,2,\ldots). Then

    y_{k^2+1} = \frac{x_{k^2+1}}{k^2+1} = \frac k{(k+1)(k^2+1)},\qquad y_{k^2} = \frac{x_{k^2}}{k^2} = \frac {k-1}{k^3},

    and you can check by multiplying out the fractions that y_{k^2+1} > y_{k^2} for all k. Thus there are infinitely many values of n for which y_{n+1}>y_n.

    That example does not quite answer the original question, because the sequence (x_n) is not strictly increasing (being constant throughout the interval k^2+1\leqslant n\leqslant (k+1)^2). However, in principle there is no difficulty in adjusting the sequence so as to make it strictly increasing. For each k, you can make x_{k^2+1} very slightly smaller, without disturbing the property that y_{k^2+1} > y_{k^2}. Then you can interpolate the values of x_n linearly for k^2+1\leqslant n\leqslant (k+1)^2 so as to ensure that the sequence increases strictly.
    that was brilliant!! thanks. it solved my question.
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  5. #5
    Senior Member abhishekkgp's Avatar
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    Re: does the sequence necessarily become decreasing??

    Quote Originally Posted by chisigma View Post
    If we define \delta_{x}(n)= x_{n+1}-x_{n} then it must be...

    a) \delta_{x}(n)>0\ \forall n

    b) \lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1 (1)

    Now, if y_{n}=\frac{x_{n}}{n} we define \delta_{y}(n)= y_{n+1}-y_{n} and in few steps we find that...

    \delta_{y}(n)= \frac{\delta_{x}(n)-y_{n}}{n+1} (2)

    Because the series in (1) converges, it must be necesserly for some \varepsilon>0 and N \delta_{x}(n)<\frac{\varepsilon}{n}\ \forall n>N and that means that \delta_{x}(n)<y_{n}\ \forall n>N so that \forall n>N is \delta_{y}(n)<0...

    Kind regards

    \chi \sigma
    thank you chisigma for your reply. It seems your answer to my question is "yes, the sequence necessarily becomes decreasing after a sufficiently large value of N." In that case i will take some time to review your solution. You may also want to read a very "close counterexample" provided by opalg.
    thanks again.
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  6. #6
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    Re: does the sequence necessarily become decreasing??

    Quote Originally Posted by chisigma View Post
     x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1 (1)

    ...

    Because the series in (1) converges, it must be necessarly for some \varepsilon>0 and N \delta_{x}(n)<\frac{\varepsilon}{n}\ \forall n>N
    That is not true. Suppose for example that a_n is always zero unless n is a perfect cube, with

    a_n = \begin{cases}1/\sqrt n = k^{-3/2} & \text{if }n=k^3, \\ 0 & \text{otherwise.} \end{cases}

    Then \textstyle\sum a_n = \textstyle\sum k^{-3/2} < \infty. But if n=k^3 then na_n = k^{3/2} \to\infty as k\to\infty. Therefore there do not exist \varepsilon>0 and N such that a_n<\frac{\varepsilon}{n} for all n>N.

    The point there, as in the OP's problem, is that a series or sequence can have bad behaviour sporadically while still being well-behaved as a whole.
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  7. #7
    MHF Contributor chisigma's Avatar
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    Re: does the sequence necessarily become decreasing??

    Quote Originally Posted by chisigma View Post
    If we define \delta_{x}(n)= x_{n+1}-x_{n} then it must be...

    a) \delta_{x}(n)>0\ \forall n

    b) \lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1 (1)

    Now, if y_{n}=\frac{x_{n}}{n} we define \delta_{y}(n)= y_{n+1}-y_{n} and in few steps we find that...

    \delta_{y}(n)= \frac{\delta_{x}(n)-y_{n}}{n+1} (2)

    Because the series in (1) converges, it must be necesserly for some \varepsilon>0 and N \delta_{x}(n)<\frac{\varepsilon}{n}\ \forall n>N and that means that \delta_{x}(n)<y_{n}\ \forall n>N so that \forall n>N is \delta_{y}(n)<0...
    May be it is necessary some more explanation of me so that I reported my post. The basic starting points are...

    a) \delta_{x}(n)>0\ \forall n

    b) \lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1 (1)

    ... and the a) means that 'punctured infinite sums' like \sum_{n} \delta_{n} where...

    \delta_{n}=\begin{cases} \frac{1}{\sqrt{n}} &\text{if}\ n=k^{3}\\ 0 &\text{otherwise}\end{cases} (2)

    ... aren't adequate counterexamples...

    Kind regards

    \chi \sigma
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  8. #8
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    Re: does the sequence necessarily become decreasing??

    Quote Originally Posted by chisigma View Post
    May be it is necessary some more explanation of me so that I reported my post. The basic starting points are...

    a) \delta_{x}(n)>0\ \forall n

    b) \lim_{n \rightarrow \infty} x_{n}= x_{1} + \sum_{k=1}^{\infty} \delta_{x}(k)= x_{0}\le 1 (1)

    ... and the a) means that 'punctured infinite sums' like \sum_{n} \delta_{n} where...

    \delta_{n}=\begin{cases} \frac{1}{\sqrt{n}} &\text{if}\ n=k^{3}\\ 0 &\text{otherwise}\end{cases} (2)

    ... aren't adequate counterexamples...
    If you want a more "adequate" counterexample, you could take

    \delta_{n}=\begin{cases} \frac{1}{\sqrt{n}} &\text{if}\ n=k^{3}, \\ \frac1{n^2}&\text{otherwise}.\end{cases}

    The fact is that the convergence of the series \textstyle\sum\delta_n does not imply that there exist \varepsilon>0 and N such that \delta_n<\frac{\varepsilon}{n} for all n>N.
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