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Thread: Limit 5

  1. #1
    Junior Member
    Dec 2009

    Limit 5

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

    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:$\displaystyle -\frac 12$)
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  2. #2
    MHF Contributor

    May 2008
    Quote Originally Posted by ns1954 View Post

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

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

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

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

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

    And anestly, I know very little about Riman's zeta-function.
    Last edited by ns1954; Jan 23rd 2010 at 06:29 PM.
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