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Thread: question about p-series

  1. #1
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    question about p-series

    The question is to find the set of p such that the series
    question about p-series-codecogseqn-2-.gif
    converge.
    I have no idea to deal with the ln() inside the sigma.
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  2. #2
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    Re: question about p-series

    $\displaystyle \sum_{n\ge 0} \ln\left(1+\dfrac{1}{n^p}\right) = \sum_{n\ge 0}\left[\sum_{k\ge 1} \dfrac{(-1)^{k+1}}{k}(n^{-kp})\right]$. Can you take it from there?
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  3. #3
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    Re: question about p-series

    To SlipEternal : I get no idea to start from your equation, though i got another idea.
    By Series Comparison Test(and L'Hospital's Rule):
    question about p-series-codecogseqn-3-.gif
    That means the series perform as np series while n goes to ∞, which converge in (1,∞)
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  4. #4
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    Re: question about p-series

    Quote Originally Posted by alter027 View Post
    To SlipEternal : I get no idea to start from your equation, though i got another idea.
    By Series Comparison Test(and L'Hospital's Rule):
    Click image for larger version. 

Name:	CodeCogsEqn (3).gif 
Views:	4 
Size:	1.8 KB 
ID:	37770
    That means the series perform as np series while n goes to ∞, which converge in (1,∞)
    What do you mean in $(1,\infty)$? Do you mean it converges if $p \in (1,\infty)$? If that is what you meant, then yes.
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  5. #5
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    Re: question about p-series

    Quote Originally Posted by alter027 View Post
    To SlipEternal : I get no idea to start from your equation, though i got another idea.
    By Series Comparison Test(and L'Hospital's Rule):
    Click image for larger version. 

Name:	CodeCogsEqn (3).gif 
Views:	4 
Size:	1.8 KB 
ID:	37770
    That means the series perform as np series while n goes to ∞, which converge in (1,∞)
    By the way, from my equation, the place to start is by swapping the sigmas:

    $\begin{align*}\displaystyle \sum_{n\ge 1} \ln\left(1+\dfrac{1}{n^p}\right) & = \sum_{n\ge 1}\left[\sum_{k\ge 1} \dfrac{(-1)^{k+1}}{k}\dfrac{1}{n^{kp}}\right] \\ & = \sum_{k\ge 1}\left[\dfrac{(-1)^{k+1}}{k}\sum_{n\ge 1}\dfrac{1}{(n^k)^p}\right]\end{align*}$

    Now, you have an alternating series. By the alternating series test, it will converge if every term converges and tends towards zero. Consider the inner summation. If $k=1$, this is a p-series. It converges when $p>1$. So, that is the initial limitation. Next, for any $k>1$, $\dfrac{1}{(n^k)^p} < \dfrac{1}{n^p}$, so by comparison test, if $p>1$, you have convergence for all $k$. Since $\displaystyle \sum_{n\ge 1}\dfrac{1}{n^{kp}}$ is decreasing, $\displaystyle \dfrac{1}{k}\sum_{n\ge 1} \dfrac{1}{n^{kp}} \to 0$ as $k \to \infty$. This proves that it is convergent for any $p>1$.
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