Zeta Function Convergence in Interval (0,1): Does It Tend to Zero?

[Charles49]

Inside the interval $$(0, 1)$$ does the function
$$
\frac{-\zeta(s)}{s}
$$
tend to zero uniformly as $$\Im(s)$$ increases?

 

[DonAntonio]

Inside the interval $$(0, 1)$$ does the function
$$
\frac{-\zeta(s)}{s}
$$
tend to zero uniformly as $$\Im(s)$$ increases?

You must explain yourself a little more (or better): what do you mean by "in the interval \,\,(0,1)\,\," and then "as \,\,\Im(x)\to\infty\,\,"? Do you mean the

real part of z remains in the unit interval while its imaginary part tends to zero or what?

DonAntonio

 

[Charles49]

Sorry for not being clear. What I meant was

Suppose $$x\in(0,1).$$ Is this true
$$
\lim_{y\rightarrow\infty}\bigg\lvert\frac{\zeta(x+iy)}{x+iy}\bigg\rvert=0?
$$

 

[jackmell]

The zeta function is analytic in that region. Why wouldn't the limit go to zero then?

 

[Charles49]

Ok. I see. It make sense. Thanks Jackmell

 

[jackmell]

I may not have that right. For example, the functioon f(z)=z is also analytic in the region yet the limit
\lim_{y\to\infty} \left|\frac{f(z)}{z}\right|
is not zero.

Here's what I suggest: I would just flat-out check it numerically first to see if it looks like it's indeed going to zero keeping in mind numeric trends may not reflect the actual limit. Now, if it looks to be the case, then I'd try and put a bound on zeta, for example, if zeta along that line has order such as O(\sqrt{z}), then the limit is obvously zero, in fact any order less than z would mean the limit is zero. Thus, if the numeric results suggests its going to zero, then I'd try and determine it's order there.

 

[Petek]

Do you know (or can you prove) the following formula:

\zeta(s) = \frac{1}{s-1} + 1 - s \int_{1}^{\infty} \frac{u - \lfloor{u}\rfloor}{u^{s+1}} du

which is valid for all s \neq 1 and for Re(s) > 0? I think that this formula can be used to prove that your limit goes to 0.

If this hint isn't helping, can you provide some more context? Such as, what have you already covered about the zeta function? Are you using a specific text?

 

[DonAntonio]

Inside the interval $$(0, 1)$$ does the function
$$
\frac{-\zeta(s)}{s}
$$
tend to zero uniformly as $$\Im(s)$$ increases?

For complex \,\,s\,,\,\,Re(s)>0\,,\,\,s\neq 1\,\, , we have the analytic cont. of the zeta function \zeta(s)=\frac{\eta(s)}{1-2^{1-s}}\,\,,\,\,\eta(s):=\sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^s}\Longrightarrow -\frac{\zeta(s)}{s}=\frac{\eta(s)}{s(2^{1-s}-1)}
Now, \,\,|2^{1-s}|=\left|e^{(1-s)Log(2)}\right|=e^{(1-Re(s))Log(2)}=2^{(1-Re(s))}\,\,\left|\eta(s)\right|\leq\sum_{n=1}^\infty\frac{1}{n^{Re(s)}} so we have that \left|-\frac{\zeta(s)}{s}\right|\leq\frac{1}{|s|(2^{1-Re(s)}-1)}\sum_{n=1}^\infty\frac{1}{n^{Re(s)}}
As \,\,0< Re(s)< 1\,\, , the above series is divergent for any constant value of s, but the value

\,\,|s|\,\, in the denominator diverges to infinity as well, so it is a matter of comparison between both divergent factors

in that fraction...

I guess you also must be aware of the fact that \,\,-\frac{\zeta(s)}{s}=\int_0^\infty\left\{\frac{1}{t}\right\}t^{s-1}dt\,\, precisely for \,\,0<Re(s)<1\,\, , so perhaps it'd be easier (?) to evaluate the limit according to this integral since as the imaginary

part of s increases to infinity the module of that power in the integral remains constant...

DonAntonio

 

[Charles49]

Sorry for responding late. I had finals this week and I was buried in my books. I am very happy to see all these responses.

jackmell, I plotted the function and it is decreasing.

Petek and DonAntonio, thanks for the hints. I entered the following in Mathematica:

In[3]:= Limit[1/x*Sum[1/n^x, {n, 1, \[Infinity]}], x -> \[Infinity]]

Out[3]= 0

What does this mean?

 

[jackmell]

Ok, I did what Pete suggested: I divided that expression by s, converted the integral into an infinite sum, and then took the antiderivative, and then took the limit. But I didn't do it rigorously.

Have you tried that approach?

 

[Charles49]

Jackmell, Pete's suggestion works for Re(s)>0.