The Riemann
Zeta function often enters through powerful complex-analytic machinery, making
its deepest properties seem inaccessible.
Here we take
a different route, developing asymptotic bounds for 
and 
using direct elementary
estimates.
In this post we proved that
for a sufficiently large 
and a chosen constant 
, provided that 
.
The proof begins with the formula
which can be shown to
be valid for 
(see Analytic
Continuation of the Zeta function).
Note: As explained
at the end of Ramanujan’s Master
Theorem, the quantity 
arises naturally in the
analytic continuation of the zeta function. In fact, one obtains the integral
representation
Differentiating
we get
Each term of 
satisfies 
. As 
grows, we choose 
such that 
. For any 
, 
and for any positive constant 
we have 
.
If then
That is, 
. Hence,
and
because in Example 2 (see A
Path from Numbers to Asymptotic Summation), we calculated the growth of the sum 
, showing that
where 
.
This completes Step 1. From the
identity
we are left with four remaining terms,
each of which must be estimated separately.
■
Step 2: The integral 
We have
Integrating by parts
Thus
Let
so
because
hence
Thus, for sufficiently large ,
However, to get
we need 
but for 
, 
not 
So, instead, we are applying the binomial expression: 
) giving
Since 
hence
for some constants 
. Hence
Thus,
■
Step 3: The term 
because for , 
Thus,
■
Step 4: The term 
From the previous step, for ,
so
■
Step 5: The term 
■
Putting
everything together
Let 
That is,
giving
Thus,
uniformly in the region
for a sufficiently large 
and a chosen constant 
.
■
Logarithmic expansion of the Zeta Function and Behavior Near 
As we know
We have 
, so we can use the power series
for 
, i.e.,
and
Combining we get
where 
is the Prime Zeta function.
Moreover, for 
(see: The Riemann Hypothesis Revealed, Zeta’s values near one)
hence
and since 
near 
, here 
, we have
■
An Upper Bound for 
Clearly, from
we get
and hence
Since 
and
implying
Thus,
or
Now
hence
Therefore, for 
where 
(simple pole at 
implies cube has triple
pole at ) and 
.
Then
and
Recall (see,
for example this),
Thus,
■
From a lower bound on 
to a Zero-Free Region
Observe that
Two cases: and 
Case 1:
We have
so
(because 
)
Consider the expression
as a function of 
:
Take the derivative with respect to 
:
gives:
which is a global minimum.
Thus,
for all .
If 
, for 
such that
then 
. That is,
where 
and 
.
Thus,
Case 2: For 
hence (since 
)
with
or
Let 
.
We have the region
where 
has no zeros (
and
Now since in the region 
, choosing 
gives
Additionally,
Epilogue
The
remarkable lesson is not merely that the estimates work, but how far elementary
reasoning can reach.
Beginning
with a simple continuation formula and careful asymptotic bookkeeping, we
arrive at conclusions that touch the frontier separating primes from zeros.
Sometimes
the shortest route toward deep territory is not a new theorem, but a clearer
view of familiar tools.
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