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Thread: Sum to infinity of log n/n! series

  1. #1
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    Sum to infinity of log n/n! series

    Hello,

    I require the value of the following sum for some complexity analysis. I would be grateful if someone can help me out. Even a valid upper bound is okay.

    $\displaystyle \sum_{i=2}^{\infty} \frac{\log i}{(i-2)!}$

    Pardon me if I have done/said anything wrong.

    Abhiram.
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    I'm assuming you're using log base e: ln. Well, $\displaystyle \ln(i)\le i\;\forall\,i> 0$. Hence,

    $\displaystyle
    \sum_{i=2}^{\infty} \frac{\log(i)}{(i-2)!}\le\sum_{i=2}^{\infty} \frac{i}{(i-2)!}=3e.
    $
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  3. #3
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    Oh well, yes. That is a valid bound. I was hoping for a tighter one. And it is $\displaystyle \log_2$. Empirically I find that it converges to 4.10.
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  4. #4
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    Ok, instead of using $\displaystyle \log_{2}(i)\le i$, we can use a different slope line through the origin. You can show that the line

    $\displaystyle y=\frac{1}{2^{1/\ln(2)}\ln(2)}\,x\ge\log_{2}(x)\,\forall\,x>0$. This is the line through the origin that just touches the $\displaystyle \log_{2}(x)$ graph. The bound is now

    $\displaystyle 3e\,\frac{1}{2^{1/\ln(2)}\ln(2)}\le 4.3281$.

    You could probably do even better if you were consider lines through the point (0,5) that are tangent to the graph of $\displaystyle \log_{2}(x)$.
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  5. #5
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    Thumbs up

    That helps. Thanks. I just wanted to know whether there exists an absolute sum. I guess not.
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  6. #6
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    Quote Originally Posted by abhiramn View Post
    Hello,

    I require the value of the following sum for some complexity analysis. I would be grateful if someone can help me out. Even a valid upper bound is okay.

    $\displaystyle \sum_{i=2}^{\infty} \frac{\log i}{(i-2)!}$

    Pardon me if I have done/said anything wrong.

    Abhiram.
    This thing converges so fast that you do not need many terms to get a good estimate:

    $\displaystyle \sum_{i=2}^{10} \frac{\log( i)}{(i-2)!}=2.847400..$

    and it is fairly easy to bound the error on this

    $\displaystyle
    \text{error}=\sum_{i=11}^{\infty} \frac{\log (i)}{(i-2)!} < \frac{\log(11)}{9}\sum_{i=11}^{\infty} \frac{1}{(i-3)!}$ $\displaystyle =\frac{\log(11)}{9}\left(e-\sum_{i=3}^{10} \frac{1}{(i-3)!}\right) \approx 7.4229 \times 10^{-6}
    $


    You can also obtain a lower bound on the error by observing that

    $\displaystyle \text{error}=\sum_{i=11}^{\infty} \frac{\log (i)}{(i-2)!} > \log(11)\sum_{i=11}^{\infty} \frac{1}{(i-2)!}\approx 7.3342 \times 10^{-6}$

    CB
    Last edited by CaptainBlack; Jun 10th 2010 at 11:19 PM.
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  7. #7
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    CB: are you using $\displaystyle \log_{2}(i)$? Or is this a lower bound? The number you came up with seems a bit low to me, at least given the numerical approximation to the sum that Mathematica gives me.
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  8. #8
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    Quote Originally Posted by Ackbeet View Post
    CB: are you using $\displaystyle \log_{2}(i)$? Or is this a lower bound? The number you came up with seems a bit low to me, at least given the numerical approximation to the sum that Mathematica gives me.
    No $\displaystyle \log_e$ but if you intended $\displaystyle \log_2$ in the original there is a simple scale factor to get that from what is in my post, since $\displaystyle \log_e(x)=\log_2(x)/\log_2(e)$. So if base 2 was intended multiply through by $\displaystyle \log_2(e)$

    CB
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  9. #9
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    Got it. Yeah, in post #3, the OP-er mentions it's log base 2.
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