I am delighted to launch the first episode of Season 2,
drawing directly from three paragraphs of my book, The Riemann Hypothesis Revealed. This new season continues our journey through the elegant
landscape of complex analysis and analytic number theory, offering readers both
foundational insights and advanced perspectives.
In my book, we explore how entire (complex analytic)
functions with prescribed zeros can be represented as infinite products. This
leads us naturally to the celebrated Weierstrass factorization theorem,
which reveals how the structure of an entire function is intimately tied to the
distribution of its zeros. Along the way, we discuss the convergence of
infinite products, the role of canonical products, and the deep connections to
analytic number theory.
Briefly, the reader will discover the following sections:
Infinite Products
This section discusses how entire (complex analytic)
functions with prescribed zeros can be represented as infinite products. If a
function 
is analytic everywhere (entire) and
has simple zeros at points 
, then locally around these zeros, can be written as 
, where 
is analytic and nonzero near the
zeros. As the neighborhood expands to include all zeros, the logarithmic
derivative 
can be represented as a sum over the
zeros, and, under suitable convergence conditions, itself can be written as an infinite
product involving its zeros and certain correction factors (to ensure
convergence). This leads to the Weierstrass factorization theorem, which
states that an entire function can be written as a product over its zeros,
possibly multiplied by an exponential of an entire function.
On the Convergence of Infinite
Products
This section explains the conditions
under which infinite products of complex numbers (or functions) converge. An
infinite product 
converges absolutely if 
and the series 
converges absolutely. For analytic functions,
if each factor is close to 1 in a certain region, then the product converges
uniformly and absolutely in that region. The section also discusses Blaschke
products (infinite products over the unit disk) and the convergence
criteria for such products. The logarithmic derivative of an infinite product
is the sum of the logarithmic derivatives of its factors, provided the product
converges uniformly. The section further introduces canonical products—infinite
products with specific correction factors (elementary factors) chosen to ensure
convergence, especially when the zeros do not grow fast enough. The Weierstrass
factorization theorem is revisited, stating that any entire function can be
written as a product over its zeros, with suitable correction factors and an
exponential of an entire function.
Hadamard’s Theorem
This section builds on the previous
discussions to introduce Hadamard’s factorization theorem. It states
that if an entire function has zeros 
and is of finite order 
, then can be written as
where 
is a polynomial (of degree at most the genus),
is the multiplicity of zero at the origin, and
are elementary factors chosen to ensure
convergence. The order of an entire function describes its growth
rate at infinity, and the genus 
is related to the minimal correction needed
for convergence of the product over zeros. Hadamard’s theorem provides bounds
on the order and genus: 
. If the order is not an integer, the
function must have infinitely many zeros. The theorem is crucial in analytic
number theory, especially in the study of the Riemann zeta function and its
zeros.
In summary:
- Infinite products allow entire functions to be
expressed in terms of their zeros, with convergence ensured by suitable
correction factors.
- The convergence of such products is tied to
the growth of the zeros and the function itself.
- Hadamard’s theorem gives a precise
factorization for entire functions of finite order, relating the
function’s growth, its zeros, and the structure of the infinite product
representation.
The following exercises are
designed to reinforce the concepts introduced in the book, such as the
convergence of infinite products, the relationship between products and sums,
and the subtleties of analytic continuation.
Let’s turn to the exercises and
see these ideas in action.
In the classic text “A Collection
of Problems on Complex Analysis” (Dover Publications, 1965) by L.I. Volkovyskii,
G.L. Lunts, and I.G. Aramanovich, I came across the following exercise:
Exercise 1
Prove that, as is customary, if we restrict the arguments of the
sequence 
to 
, then the convergence or divergence of the infinite product 
is equivalent to the convergence or
divergence of the series 
.
Let’s do this carefully.
We assume:
for all n,
we use the principal branch of the logarithm, so ,
and (as is standard for infinite
products) we are interested in whether
converges (to a nonzero limit) or diverges, and we compare this
to
Partial products and partial sums
Define the partial products
and the partial sums
where log is the principal branch.
Whenever the principal branch is consistently defined
along the sequence , we have
This argument is too vague.
The problem is that
is false in general for the principal logarithm. It is
true when no branch cut is crossed.
More precisely, for the principal branch one has
for some integer k.
So to be allowed to write
we must assume something like: for every N, the straight multiplicative accumulation of the arguments does not cross the negative real axis.
A standard sufficient hypothesis is:
This is exactly what “consistently defined” means.
Hence if we assume that for every N, the partial product 
avoids 
then
So after removing the ambiguity:
This identity is the core link between the product and the sum.
If 
converges, then 
converges
Assume the series
converges in 
. Then the sequence of partial sums 
converges to some limit 
.
Since the exponential function is continuous,
So the infinite product converges (to the nonzero limit P).
If converges (to a nonzero
limit), then converges
Now assume the infinite product
converges to some nonzero limit 
. Then the partial products form a Cauchy sequence
and
Because , for all sufficiently large N, stays in some compact
subset of 
that does not cross the
branch cut of the principal logarithm. Thus, for all large N, 
is well-defined and
continuous in N, and we can write
Strictly speaking we need:
for all sufficiently large N, 
, so that the principal logarithm is
single-valued and continuous in a neighborhood of the tail of the sequence.
Having established the right condition, since 
and log is continuous on 
, we may conclude
Hence the series converges (its partial
sums converge to 
)
Divergence
if 
diverges in , then the sequence does not converge, so cannot converge to a
nonzero limit. The product either diverges or converges to 0 (in which case 
).
The reason is not simply “because does not converge”. In
general, one can have a sequence which does not converge
in but for which 
does converge (for
example by adding integer multiples of 
).
What saves the argument is that: we are fixing the principal
logarithms of the individual factors, and we require the identity .
Conversely,
if the product does not converge to a nonzero limit, then does not converge in , so cannot converge in .
Thus, under the principal-branch restriction (so that 
is consistently defined), we have:
converges (to a nonzero
limit) 
converges
assuming that the partial products
avoid the branch cut .
And in the extended sense (allowing convergence of the product
to 
), divergence of one corresponds to divergence of the other.
One more missing classical hypothesis
There is a standard extra assumption which is usually stated
explicitly in textbooks:
converges to a non-zero
limit if and only if converges and 
.
In fact, from convergence of we automatically get , but in the reverse direction authors often state it to avoid
pathological formulations.
Exercise 2
Now we have to check whether the assertion of the preceding
problem is still true if we assume that: 
or 
, 
This is exactly where the branch-cut geometry starts to bite.
We’re now changing the way we choose arguments of 
, hence changing the branch of log. The previous result relied
crucially on using a single continuous branch of log along the sequence of
partial products.
Let’s look at the two new cases.
Case 1:
Here we’re using arguments in 
. That’s the principal argument shifted: instead of 
, we’ve rotated the cut so that branch cut is now along the
positive real axis.
The problem: as n grows the partial products
may cross the positive real axis
infinitely often. Each time you cross the branch cut, the value of jumps by . So:
- We can still define
with , - but
is no longer guaranteed to equal
in a way that is
continuous in N. - The
identity itself can fail, not only continuity because with this branch,
for some integers 
, and these integers can change infinitely
often if the partial products cross the cut infinitely often. This is the real
obstruction.
So the key identity
can fail to behave nicely as N increases, because of branch-cut jumps. That breaks the equivalence:
- It
is possible for while on the chosen path
differs from by varying multiples of ,
- or
for to converge in this
branch while the product behaves differently under another branch.
Conclusion for this case: The clean
equivalence “product converges sum of logs converges” is no longer
guaranteed if you use .
Case 2: ,
Same story, different rotation.
We’re now using a branch of log whose
cut lies along the ray 
. Again, the crucial question is:
Do the partial products stay in a region that does not cross this
branch cut for all large N?
If they do, then for all
sufficiently large N, is a continuous function of N on that branch,
and you recover
in a stable way. Then the same
equivalence as before holds.
If they don’t, i.e. if crosses the branch cut infinitely often, you
again get jumps of in , and the neat equivalence between
convergence of and can fail.
So:
The assertion from the previous problem is not automatically
true for arbitrary choices of argument intervals like or .
It remains true only if the
chosen branch of log is such that the partial products eventually stay in a
simply connected region avoiding the branch cut, so that is single–valued and continuous along
the sequence 
. In fact, it is actually enough that
(branch cut) for all large N. and
that all those points lie in the same connected component of the cut plane.
In other words: the equivalence is a
statement about choosing a consistent branch of log
along the path of partial products, not about
any arbitrary way of assigning arguments.
__________
Moving beyond these pathological
cases, we have the following result:
Exercise 3
Prove that for the absolute
convergence of the infinite product
that is, for the absolute convergence
of the series
it is necessary and sufficient that
the series
should converge absolutely.
In my book, I prove that absolute convergence of
the series 
ensures the absolute convergence of the
infinite product 
. This result serves
as a motivating principle for what
follows. That’s why I named it The Riemann Hypothesis Revealed and not Complex
Analysis. When writing about the Riemann Hypothesis and striving for minimal
compromises in exposition, certain technical details are sometimes omitted for
focus. My readers will know by now that I don’t follow academic conventions.
The reason is simple: my priority is to make mathematics transparent and
accessible, rather than adhering to traditional formalities that can obscure
the essential ideas.
However, through this blog, we have
the opportunity to revisit, validate, and enrich the story. Here, we can
explore the technical nuances that are sometimes omitted in the book for the
sake of clarity, offering readers a more comprehensive perspective. By
expanding on these foundational principles, we not only reinforce the
motivation behind key results, but also provide a deeper understanding of their
role in analytic number theory and the journey toward the Riemann Hypothesis.
Let’s see both directions now.
convergent 
product absolutely convergent
Assume
Then 
. So there exists N such that for
all 
,
For 
, the Taylor series
converges absolutely and we have
uniform estimate
for some constant 
and all .
Apply this with 
for to get
The finite initial segment 
is irrelevant for absolute convergence, so
converges absolutely. By definition
the product converges absolutely.
Product absolutely convergent convergent
Now assume the product converges
absolutely, i.e.
converges absolutely.
Then in particular 
, hence 
, so . For n large, 
, and again we can use Taylor
expansion.
For small z, we have the
reverse-type estimate
for some constant and all . This follows from 
and continuity near 0.
Thus, for large n,
Since 
converges, the comparison test gives
So absolute convergence of the
product is equivalent to absolute convergence of 
.
Exercise 4
The infinite products and 
converge. Investigate the convergence of the
infinite products
Let’s assume “converge” in the usual
sense for infinite products of complex numbers: converges means , and converges (similarly for 
).
So for large n 
.
Product 
Here 
also 
so 
a sum of two convergent series. Hence 
converges.
Similar for 
.
Product 
Now 
so the product diverges (blows up in modulus).
Product 
Here 
. Therefore, it is impossible to have
so the product always diverges.
Exercise 5
Investigate the convergence of the
following products
(i)
, 
(ii)
(iii)
For (i), it is natural to write the factors as pairs.
Hence,
However, if we set
then and 
.
Thus,
We have 
, i.e., converges (alternating
harmonic) and 
converges. So 
converges, hence the product converges to a
nonzero limit.
But 
diverges, so the product is not absolutely convergent
only conditionally.
Conclusion: convergent
but not absolutely.
For (ii), we have convergence
for 
and divergence for 
. Reader’s exercise
For (iii), notice that
and
diverges like a harmonic series.
Reader’s exercise.
Epilogue
Season 2 Episode 1 has taken us deep
into the heart of complex analysis, exploring the elegant machinery of infinite
products. These are not just technical results—they are the keys that unlock
the hidden architecture of entire functions and illuminate the path toward some
of mathematics’ most celebrated mysteries.
If you found this episode insightful,
I invite you to take the next step and invest in your mathematical growth by
purchasing my book, The Riemann Hypothesis Revealed. This book is more
than just a collection of theorems—it’s a carefully crafted guide designed to
make advanced concepts accessible, transparent, and inspiring. Whether you’re a
student, researcher, or lifelong learner, you’ll find that this investment pays
dividends in clarity, confidence, and a deeper understanding of the
mathematical universe.
Don’t just read about
mathematics—experience it fully. Make The Riemann Hypothesis Revealed
your next investment in knowledge and discovery.
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