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 $\displaystyle \frac{\mathrm{d} }{\mathrm{d} x}a^x=(\log a)a^x$ 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 
Banned Camp Joined: Mar 2015 From: New Jersey Posts: 1,720 Thanks: 125  
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?

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.


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derivative, function, infinity, involving, limit, zeta, zeta function 
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