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Thread: convergent series..

  1. #1
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    convergent series..

    i have a sequence of function and i cannot find the correct bound for it. can someone help me out here? thanks in advanced!

    Let a>1; x any real number..

    1/(1+|x+k|)^a < M(sub)k such that the series M(sub)k, k= -infinity to +infinity is convergent.
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  2. #2
    Super Member girdav's Avatar
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    You can study the function $\displaystyle f_k$ deifned by $\displaystyle f_k\left(x\right) = \frac 1{\left|x+k\right|^a}$ ($\displaystyle x$ can't be an integer).
    But it seem that it's not a bounded function: what is $\displaystyle \lim_{x\to -k}f_k\left(x\right)$
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  3. #3
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    thanks for the reply.

    i actually want M(sub)k to be a sequence of real numbers, so it must be independent of x.
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  4. #4
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    Quote Originally Posted by minesweeper26 View Post
    i have a sequence of function and i cannot find the correct bound for it. can someone help me out here? thanks in advanced!

    Let a>1; x any real number..

    1/(1+|x+k|)^a < M(sub)k such that the series M(sub)k, k= -infinity to +infinity is convergent.
    I don't think you can have such $\displaystyle M_k$ because $\displaystyle f_k(x)=\frac{1}{1+|x+k|^a} \leq 1$ but actually attains this vaue at $\displaystyle -k$ so the M-test can't be applied on the whole line (Maybe in some interval you could).

    Edit: For example for $\displaystyle k\geq 1$ we have $\displaystyle f_k(x)\leq \frac{1}{k^a}$ on $\displaystyle (0,\infty)$ and $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^a} <\infty$. Similarly you can apply the M-test to $\displaystyle f_k$ for $\displaystyle k\leq -1$ on $\displaystyle (-\infty , 0)$.
    Last edited by Jose27; Jan 30th 2010 at 08:18 PM.
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  5. #5
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    Quote Originally Posted by Jose27 View Post
    I don't think you can have such $\displaystyle M_k$ because $\displaystyle f_k(x)=\frac{1}{1+|x+k|^a} \leq 1$ but actually attains this vaue at $\displaystyle -k$ so the M-test can't be applied on the whole line (Maybe in some interval you could).

    Edit: For example for $\displaystyle k\geq 1$ we have $\displaystyle f_k(x)\leq \frac{1}{k^a}$ on $\displaystyle (0,\infty)$ and $\displaystyle \sum_{k=1}^{\infty} \frac{1}{k^a} <\infty$. Similarly you can apply the M-test to $\displaystyle f_k$ for $\displaystyle k\leq -1$ on $\displaystyle (-\infty , 0)$.
    inequality is not true for all x.. note that x can have a negative value.. hence, f(sub)k may not be less than 1/k^a..

    another thing, a is the exponent of the whole denominator not just the absolute value..

    thanks anyway..

    EDIT: i mean, how will you bound f(sub)k on (-inf,0) for k>0?
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