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Thread: Limit 5

  1. #1
    Junior Member
    Dec 2009

    Limit 5

    <br />
\lim \limits_ {x\to {0^+}}\sum_{n=1}^{\infty}\frac {(-1)^n}{n^x}<br />

    Hint: Define suitable sequence, use Abel's transformation to transform general term of series and Dirichle's theorem about uniform convergence of functional series. (Answer: -\frac 12)
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  2. #2
    MHF Contributor

    May 2008
    Quote Originally Posted by ns1954 View Post

    Find \lim \limits_ {x\to {0^+}}\sum_{n=1}^{\infty}\frac {(-1)^n}{n^x}<br />
    it's a trivial result of \sum_{n=1}^{\infty}\frac {(-1)^n}{n^x}=(2^{1-x}-1)\zeta(x) that the limit is equal to -1/2.
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  3. #3
    Junior Member
    Dec 2009
    I offer this direct proff. Function f(x)=\sum_{n=1}^{\infty}\frac{(-1)^n}{n^x} is defined for x>0, by Dirichle criteria. Let a_n=-\frac12+ \sum_{k=0}^n(-1)^k=\frac {(-1)^n}2
    Now, a_n-a_{n-1}=(-1)^n for n\ge 1, and
    \left |\sum_{k=1}^n a_k\right |<1
    Let x>0. By Abel's transformation, we have

    f(x)=\sum_{n=1}^{\infty} \frac {a_n-a_{n-1}}{n^x}=<br />
-a_0+ \sum_{n=1}^{\infty}a_n[\underbrace{\frac1{n^x}-\frac1{(n+1)^x}}_{b_n(x)}]=<br />

    it's only neccessary to prove that series for g(x) is uniform convergent for x\ge 0
    If b_n (x)=\frac1{n^x}-\frac1{(n+1)^x}=\frac 1{n^x}\left[1-\frac1{(1+\frac1n)^x}\right]
    it's easy to prove that
    b_n(x)\rightrightarrows 0 and it's decreasing.
    For x\in [0,1] we have

    <br />
|b_n(x)|\leq 1-\frac1{(1+\frac1n)^x}\leq 1-\frac1{1+\frac1n}=\frac1{n+1}\ < \frac1n<br />

    and for x>1,\ |b_n(x)|\leq 1/{n^x}<1/n, so that
    \sup_{x\ge 0}|b_n(x)|<\frac1n

    And anestly, I know very little about Riman's zeta-function.
    Last edited by ns1954; Jan 23rd 2010 at 07:29 PM.
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