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November 8th, 2016, 07:00 AM   #1
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Unhappy Limit involving derivative of zeta function

Hello. I would like to know what is the value of the derivative of 1/ (zeta(x)) as x tends to 1:

$\displaystyle \displaystyle \lim_{x \to 1} \frac{d\frac{1}{\zeta(x)}}{dx}
$

As you get an indetermination of the type $\displaystyle 0\infty $, what I have tried so far is to apply L'Hôpital's rule, but with no success.

Thank you.
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November 8th, 2016, 08:40 AM   #2
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What is your definition of the zeta function? Can't find it.
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November 9th, 2016, 06:47 AM   #3
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$\displaystyle \zeta(p)=\sum_{n=1}^{\infty}\frac{1}{n^{p}}=\sum_{ n=1}^{\infty}(\frac{1}{n})^p$, p>1

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} p}\zeta(p)=\sum_{n=1}^{\infty}\log(\frac{1}{n}) \frac{1}{n^{p}}=-\sum_{n=1}^{\infty}\log(n)\frac{1}{n^{p}}$

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} p}\frac{1}{\zeta(p))}= -\frac{1}{\zeta(p)^{2}}\frac{\mathrm{d} }{\mathrm{d} p}\zeta(p)=\frac{1}{\zeta(p)^{2}} \sum_{n=1}^{\infty}\log(n)\frac{1}{n^{p}}$

$\displaystyle \lim_{p\rightarrow 1}\frac{\mathrm{d} }{\mathrm{d} p}\frac{1}{\zeta(p))}=0\lim_{p\rightarrow 1}\sum_{n=1}^{\infty}\log(n)\frac{1}{n^{p}}=?$


Ref
$\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}a^x=(\log a)a^x$
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Last edited by skipjack; November 9th, 2016 at 11:37 PM.
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November 9th, 2016, 11:18 AM   #4
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Ref previous post.

$\displaystyle \frac{\mathrm{d} }{\mathrm{d} p}\frac{1}{\zeta(p))}= -\frac{1}{\zeta(p)^{2}}\frac{\mathrm{d} }{\mathrm{d} p}\zeta(p)=\frac{1}{\zeta(p)^{2}} \sum_{n=1}^{\infty} \log(n)\frac{1}{n^{p}}$

$\displaystyle \lim_{p\rightarrow 1}\frac{\mathrm{d} }{\mathrm{d} p}\frac{1}{\zeta(p))}= \lim_{p\rightarrow 1}\sum_{n=1}^{\infty}\frac{\log(n)\frac{1}{n^{p}}} {\zeta(p)^{2}}= 0$


$\displaystyle \lim_{p\rightarrow 1}\zeta(p)=\infty$

Last edited by skipjack; November 9th, 2016 at 11:45 PM.
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November 9th, 2016, 10:09 PM   #5
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Previous post incorrect. Limit is indeterminate. Situation is similar to

$\displaystyle \lim_{k\rightarrow \infty}\frac{1}{k} \sum_{n=1}^{\infty}\frac{1}{n}$

L'Hôpitals rule, differentiating with respect to p, doesn't help.

Last edited by skipjack; November 9th, 2016 at 11:47 PM.
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November 9th, 2016, 11:49 PM   #6
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$\displaystyle \lim_{x \to 1} \frac{d\frac{1}{\zeta(x)}}{dx} = 1$
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November 10th, 2016, 06:45 AM   #7
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Quote:
Originally Posted by skipjack View Post
$\displaystyle \lim_{x \to 1} \frac{d\frac{1}{\zeta(x)}}{dx} = 1$
Thanks. How'd you get it?
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November 10th, 2016, 11:44 AM   #8
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It's easily deduced from this expansion.
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November 10th, 2016, 03:49 PM   #9
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How?
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November 14th, 2016, 09:59 AM   #10
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For x close enough to 1, it gives 1/ζ(x) as being approximately x - 1, which has derivative 1.
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