My Math Forum Limit involving derivative of zeta function

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 November 8th, 2016, 07:00 AM #1 Newbie   Joined: Nov 2016 From: Spain Posts: 1 Thanks: 0 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.
 November 8th, 2016, 08:40 AM #2 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 What is your definition of the zeta function? Can't find it.
 November 9th, 2016, 06:47 AM #3 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 $\displaystyle \zeta(p)=\sum_{n=1}^{\infty}\frac{1}{n^{p}}=\sum_{ n=1}^{\infty}(\frac{1}{n})^p$, 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 $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}a^x=(\log a)a^x$ Thanks from topsquark Last edited by skipjack; November 9th, 2016 at 11:37 PM.
 November 9th, 2016, 11:18 AM #4 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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.
 November 9th, 2016, 10:09 PM #5 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 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.
 November 9th, 2016, 11:49 PM #6 Global Moderator   Joined: Dec 2006 Posts: 20,810 Thanks: 2151 $\displaystyle \lim_{x \to 1} \frac{d\frac{1}{\zeta(x)}}{dx} = 1$
November 10th, 2016, 06:45 AM   #7
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Quote:
 Originally Posted by skipjack $\displaystyle \lim_{x \to 1} \frac{d\frac{1}{\zeta(x)}}{dx} = 1$
Thanks. How'd you get it?

 November 10th, 2016, 11:44 AM #8 Global Moderator   Joined: Dec 2006 Posts: 20,810 Thanks: 2151 It's easily deduced from this expansion.
 November 10th, 2016, 03:49 PM #9 Banned Camp   Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125 How? Thanks from topsquark
 November 14th, 2016, 09:59 AM #10 Global Moderator   Joined: Dec 2006 Posts: 20,810 Thanks: 2151 For x close enough to 1, it gives 1/ζ(x) as being approximately x - 1, which has derivative 1. Thanks from topsquark

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