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Keep in mind that if two functions are divergent, that does not necessarily mean that the difference of these functions is also divergent. This is because the difference between two divergent functions can potentially converge, depending on the specific functions and their respective rates of divergence.

For example, consider the functions 1/x and Csc(x). Both of these functions are undefined or divergent at x=0. However, the function 1/x-Csc(x) is well-behaved, defined, and convergent at x=0. This is because the divergence of each individual function cancels out when their difference is taken.

Also, please note that the super symmetric equation remains valid for many non-zero values of the zeta function. For instance, assuming ζ(1/2) = 1.46035450881..., the SSE still holds. To put it differently, RSLT relates to the distribution of zeros of the zeta function, and the set of zeros of the zeta function is a subset of values that satisfy the SSE.

Later, we can leverage the remarkable property of the Transcendental Zeta Function and the super symmetric equation to prove the RSLT.

The ∑(n=1 to ∞) 1/n^(1-s*) and ∑(n=1 to ∞) 1/n^s define a function of s, which is initially defined only for s with real part greater than 1 or less than 0. However, these functions can be analytically continued to the whole complex plane (except for s = 1) by using the functional equation of the Riemann zeta function, which relates ζ(s) to ζ(1-s).

The functional equation of the Riemann zeta function is given by:

ζ(s) = 2^s π^(s-1) sin(πs/2) Γ(1-s) ζ(1-s) or ζ(1-s*)=2^(1-s*) π^(-s*) sin(πs*/2) Γ(s*) ζ(s*)

Using this functional equation, we can express the difference of the two series as:

∑(n=1 to ∞) 1/n^(1-s*) - ∑(n=1 to ∞) 1/n^s = ζ(1-s*) - ζ(s) = 2^(1-s*) π^(-s*) sin(πs*/2) Γ(s*) ζ(s*) - 2^s π^(s-1) sin(πs/2) Γ(1-s) ζ(1-s)

This expression is valid for all complex values of s except for s=1, where the functional equation has a pole. Thus, the function defined by this expression can be analytically continued to the whole complex plane (except for s=1), and the uniqueness property of analytic continuation implies that this continuation is unique.

In particular, the fact that the difference of the two series is zero for some values of s (i.e., ζ(s) = ζ(1-s*)) implies that the analytic continuation of the function defined by this equation is identically zero on the critical line Re(s) = 1/2, by the RSLT.

To put if diffrently , the super symmetric equation is a special case of the functional equation for the Riemann zeta function, given by:

ζ(s) = ζ(1-s*)

where s* is the complex conjugate of s. One interesting consequence of this equation is that if ζ(s) and ζ(1-s*) are equal to each other, then the difference between the sums of the series ∑(n=1 to ∞) 1/n^s and ∑(n=1 to ∞) 1/n^(1-s*) must converge to zero.

To prove this statement, we assume the opposite: that ζ(s) and ζ(1-s*) are equal to each other, but the difference between the sums of the series does not converge to zero. This means that there exists some ε > 0 such that for any N, there exists some n > N such that |∑_{n=1}^N \frac{1}{n^s} - ∑_{n=1}^N \frac{1}{n^{1-s*}}| > ε.

However, this contradicts the fact that ζ(s) and ζ(1-s*) are equal to each other, as the functional equation of the Riemann zeta function implies that the two series ∑(n=1 to ∞) 1/n^s and ∑(n=1 to ∞) 1/n^(1-s*) must converge to the same value.

Therefore, we conclude that if ζ(s) and ζ(1-s*) are equal to each other, then the difference between the sums of the series ∑(n=1 to ∞) 1/n^s and ∑(n=1 to ∞) 1/n^(1-s*) must converge to zero. Moreover, the converse is also true: if the difference between the sums of the series converges to zero, then ζ(s) and ζ(1-s*) must be equal to each other, according to the functional equation of the Riemann zeta function.

Welcome to our comprehensive summary of the rigorous proofs for the super symmetric equation in the critical strip, a key milestone in Riemann's Last Theorem proof. We have devoted years of extensive research, analyzing existing literature, and developing new mathematical models and techniques to tackle this challenging problem. Our tireless efforts have led to deriving a solution for the super symmetric equation, paving the way for further advancements in this field.

Our summary of the proofs offers unique insights into the intricate details and logical framework that underpin our solution, making it accessible to a broader audience of mathematicians, researchers, and enthusiasts. By providing an easy-to-follow summary of the proofs, we aim to help readers delve into the complexities of the super symmetric equation and gain a deeper understanding of the groundbreaking breakthrough that it represents.

Whether you are a seasoned mathematician or a newcomer to the field, this page serves as a valuable resource for exploring the latest advancements in mathematics and uncovering the secrets of Riemann's zeta function in the critical strip.

This proof of the super symmetric equation is exceptionally comprehensive and elegant in its simplicity. What sets this method apart is its ability to prove a fundamental property of the zeta function using only basic arithmetic operators (+, -, x, /) and straightforward set properties. As a result, this proof stands out as the most extensive and effective approach to solving the super symmetric equation. By carefully analyzing the problem and breaking it down into its essential components, the proof's creators were able to develop a logical framework that illuminates the underlying principles of the zeta function in a clear and intuitive way. This method is a testament to the power of simplicity and underscores the importance of taking a step back to examine a problem from multiple angles. Through this proof, mathematicians and enthusiasts alike can gain a deeper appreciation for the elegance and beauty of mathematical reasoning and the remarkable insights that it can provide.

This method starts with equation (12). In (12), we see that Riemann's zeta function is connected to a convergence zeta function in the critical strip via a series of back-to-back equalities. For simplicity and visualization purposes, let's call the convergence ζ(s) (origami) "bird." and Riemann's zeta function (origami) "boat." On this page, you see how we can unfold and refold Riemann's zeta function (boat) and obtain the convergence zeta function (bird). After this step, in equation (13), we see that if we assume that two birds are equal or ζ(s)=ζ(1-s*), we must say that the two boats are equal as well (and vice versa). ( We can assert equation (13) with equation (14) using set properties). Finlay in (15), we see that if we say ζ(s)-ζ(1-s*)=0 ( bird - bird =0), that means the boat-boat=0 ( and vice versa). Therefore, we conclude the symmetric equation (16).

The Abridged Riemann's Last Theorem article is a condensed version of the original article, which includes the identity theorem to establish the super symmetric equation. The proof shows that since Riemann's zeta function (s) or (g) is equivalent to the transcendental zeta function (s) or (f) for real(s) >1 (where an accumulation point exists), we can infer that Riemann's zeta function (s) or (g) is also equivalent to the transcendental zeta function (s) or (f) in the critical strip (0 < real(s) <1). A similar argument applies to Riemann's zeta function (1-s*) and the transcendental zeta function (1-s*). Consequently, we deduce that the equation, transcendental zeta function (s) - transcendental zeta function (1-s*) =0 is equivalent to Riemann's zeta function (s) - Riemann's zeta function (1-s*)=0.

The proof presented here for the super symmetric equation is remarkably concise and requires familiarity with the concept of analytic continuation. This approach stands out as the most efficient method for proving the equation, as it utilizes the power of this mathematical concept to simplify the proof into its shortest possible form. Analytic continuation is a critical tool in complex analysis that enables functions to be extended beyond their original domain, making it a fundamental concept in the study of the Riemann zeta function. With a basic understanding of this concept, one can fully appreciate the elegance and efficiency of this proof for the super symmetric equation.

Note that :The uniqueness property of analytic continuation is a fundamental concept in complex analysis that states that there can be only one possible extension of a given analytic function to a larger domain. In other words, if an analytic function defined on a domain can be extended to a larger domain in two different ways, then the two extensions must coincide on the intersection of their domains.

$$\zeta (1-\stackrel{\u203e}{s})-\text{}\zeta \left(s\right)=0$$

Proof: Let ζ(s) be the analytic continuation of the series ∑(n=1 to ∞) 1/n^s, and let ζ(1-s*) be the analytic continuation of the series ∑(n=1 to ∞) 1/n^(1-s*).

Therefore, considering the one-to-one correspondence property of analytic continuation, we can say that if ζ(s) - ζ(1-s*) = 0, then the series ∑(n=1 to ∞) 1/n^s - ∑(n=1 to ∞) 1/n^(1-s*) must also be equal to zero (and vice versa).

Assume that ζ(s) and ζ(1-s*) are equal, and consider the function f(z) defined by:

f(z) = ζ(z) - ζ(1-z)

Using the functional equation for the Riemann zeta function, we can rewrite this as:

f(z) = f(1-z)

Now, note that f(z) is an entire function (a function that is analytic everywhere on the complex plane), since it is a sum of terms of the form 1/n^z, which are analytic everywhere except at z=1. Moreover, f(z) is also an even function, since ζ(z) and ζ(1-z) are symmetric around the line Re(z) = 1/2.

Let's now consider the Laurent series of f(z) around z=1/2:

f(z) = ∑_{n=-∞}^{∞} a_n (z - 1/2)^n

Since f(z) is even, we have a_n = a_{-n} for all n. Moreover, since f(z) is entire, the Laurent series converges uniformly on any compact subset of the complex plane. In particular, it converges uniformly on the circle |z-1/2| = r for any positive real number r < 1/2.

Now, since f(z) satisfies the functional equation f(z) = f(1-z), we can use this to rewrite the Laurent series as:

f(z) = ∑_{n=-∞}^{∞} a_n (1/2 - z)^n

Using the fact that a_n = a_{-n}, we can simplify this to:

f(z) = ∑_{n=0}^{∞} a_n (1/2 - z)^n + ∑_{n=1}^{∞} a_n (z - 1/2)^n

Now, let's look at the difference between the sums of the two series ∑(n=1 to infinity) 1/n^s and ∑(n=1 to infinity) 1/n^(1-s*):

|∑_{n=1}^∞ 1/n^s - ∑_{n=1}^∞ 1/n^(1-s*)| = |f(s) - f(1-s)|

But since ζ(s) and ζ(1-s*) are equal, we have f(s) = f(1-s), so |f(s) - f(1-s)| = 0, and the difference between the two series converges to zero.

Conversely, if the difference between the two series converges to zero, then we have f(s) = f(1-s), and since f(z) is an entire function, it must satisfy this equation for all complex numbers s.

Therefore, the super symmetric equation holds true.

Assume that ζ(s) = ζ(1-s*) for some complex number s. Then, consider the function f(z) defined by:

f(z) = ζ(z) - ζ(1-z*)

By assumption, f(s) = 0. We will show that f(z) = 0 for all complex numbers z.

First, note that f(z) is an analytic function on the whole complex plane except for poles at z = 1 and z = 0. These poles do not affect the validity of the super-symmetric equation, as the two series converge absolutely for all other values of z.

Now, consider the function g(z) = f(z) / sin(πz). This function is also analytic on the whole complex plane except for poles at integer values of z. These poles are removable, as sin(πz) has zeros at those points of the same order as the poles of f(z). Thus, g(z) is actually an entire function.

We claim that g(z) is bounded. To see this, note that |sin(πz)| = e^(-Im(πz)) / 2, and so:

|g(z)| = |f(z)| / |sin(πz)| ≤ 2|f(z)| e^(Im(πz))

Since ζ(z) and ζ(1-z*) are both convergent for any fixed value of Re(z) > 1/2, we have that f(z) = ζ(z) - ζ(1-z*) is also convergent for any fixed value of Re(z) > 1/2. Thus, there exists some M > 0 such that |f(z)| ≤ M for all z with Re(z) > 1/2.

Putting it all together, we have:

|g(z)| ≤ 2M e^(Im(πz))

But e^(Im(πz)) is bounded on any horizontal strip of the form {z : a ≤ Im(z) ≤ b}, since it is periodic in Im(z) with period 2. Thus, g(z) is a bounded entire function, and by Liouville's theorem, it must be constant. In particular, g(s) = 0.

But we also have:

g(s) = f(s) / sin(πs) = 0 / sin(πs) = 0

So we conclude that f(z) / sin(πz) = g(z) = 0 for all complex numbers z. But sin(πz) is never zero, so it follows that f(z) = 0 for all complex numbers z.

Therefore, if ζ(s) = ζ(1-s*) for some complex number s, then the difference between the series ∑(n=1 to infinity) 1/n^s and ∑(n=1 to infinity) 1/n^(1-s*) must converge to zero. Conversely, if the difference between the series converges to zero, then ζ(s) = ζ(1-s*) for all complex numbers s, by the identity theorem.

Let ζ(s) be the analytic continuation of the sum of 1/n^s from n=1 to ∞, and let ζ(1-s*) be the analytic continuation of the sum of 1/n^(1-s*) from n=1 to ∞.

Assume ζ(s) = ζ(1-s*). By the identity theorem for analytic functions, ζ(s) and ζ(1-s*) are identical on a region that contains the line Re(s) = 1/2. Since the series ∑(n=1 to ∞) 1/n^s and ∑(n=1 to ∞) 1/n^(1-s*) converge absolutely for Re(s) > 1, and are analytic in this region, they must be identical in this region. Therefore, the difference between the series, i.e.,

∑(n=1 to ∞) 1/n^s - ∑(n=1 to ∞) 1/n^(1-s*),

converges to 0 as s approaches any point on the line Re(s) = 1/2.

Conversely, assume that the difference between the series converges to 0 as s approaches any point on the line Re(s) = 1/2. Let f(s) = ζ(s) - ζ(1-s*). Then f(s) is an analytic function on the half-plane Re(s) > 1/2. We want to show that f(s) = 0 for all s in this half-plane.

By the assumption, the limit of f(s) as s approaches any point on the line Re(s) = 1/2 is 0. Since the limit of an analytic function is itself analytic, f(s) is analytic on the half-plane Re(s) ≥ 1/2. By the identity theorem, if f(s) = 0 for an infinite set of points in this half-plane with a limit point, then f(s) = 0 for all s in the half-plane. Therefore, it suffices to show that the set of zeros of f(s) has a limit point in the half-plane Re(s) ≥ 1/2.

Let z = 1/2 + it, where t is a real number. Then

f(z) = ζ(z) - ζ(1-z*) = ∑(n=1 to ∞) (1/n^z - 1/n^(1-z*)).

Using the fact that |1/n^z - 1/n^(1-z*)| ≤ 2/n^(1/2), we have

|f(z)| ≤ ∑(n=1 to ∞) 2/n^(1/2) = 2ζ(1/2) < ∞.

Therefore, by the Cauchy-Schwarz inequality, we have

|f(z + ih)| ≤ √(2ζ(1/2))√(∑(n=1 to ∞) 1/n^(1 + h))√(∑(n=1 to ∞) 1/n^(1 - h)),

where h is a small positive real number. Since both ∑(n=1 to ∞) 1/n^(1 + h) and ∑(n=1 to ∞) 1/n^(1 - h) converge for h > 0, the above inequality shows that f(z) approaches 0 as h approaches 0. Therefore, the set of zeros of f(s) has a limit point at z in the half-plane Re(s) ≥ 1

The statement of the super symmetric equation is:

If ζ(s) = ζ(1-s*), then the difference between the sums of the series

∑(n=1 to infinity) 1/n^s and ∑(n=1 to infinity) 1/n^(1-s*)

converges to zero. Conversely, if the difference of the two series converges to zero, and ζ(s) - ζ(1-s*) = 0, then ζ(s) = ζ(1-s*).

Proof of the forward direction:

Assume that ζ(s) = ζ(1-s*). Then, we can write:

∑(n=1 to infinity) 1/n^s - ∑(n=1 to infinity) 1/n^(1-s*) = ζ(s) - ζ(1-s*) = 0

Therefore, the difference between the two series converges to zero.

Proof of the converse:

Assume that the difference between the two series converges to zero, and ζ(s) - ζ(1-s*) = 0. We need to show that ζ(s) = ζ(1-s*).

Let f(s) = ζ(s) - ζ(1-s*). Then, by assumption, f(s) = 0. We want to show that f(s) = 0 for all values of s.

The function f(s) is analytic in the region Re(s) > 0. By the uniqueness property of analytic continuation, if f(s) = 0 for an infinite set of points with a limit point in the region Re(s) > 0, then f(s) = 0 for all s in that region.

Let S = {s: f(s) = 0}. We want to show that S has a limit point in the region Re(s) > 0.

Suppose S has no limit point in the region Re(s) > 0. Then, S is a discrete set of points. Let s0 be the point in S with the largest real part. Then, there exists a positive real number ε such that the open disk |s-s0| < ε does not contain any other point of S.

Consider the function g(s) = f(s)/(s-s0). Then, g(s) is also analytic in the region Re(s) > 0, and g(s) = 0 for all s in S except for s0.

By the Laurent series expansion of g(s) around s0, we have:

g(s) = c/(s-s0) + h(s),

where c is a complex constant, and h(s) is an analytic function in the region Re(s) > 0.

Since g(s) = 0 for all s in S except for s0, we have c = 0. Therefore, g(s) = h(s) is an analytic function in the region Re(s) > 0.

Since S is a discrete set of points, we can choose ε small enough such that the closed disk |s-s0| ≤ ε does not contain any other point of S. Then, by the maximum modulus principle, the maximum value of |g(s

Since S is a discrete set of points, we can choose ε small enough such that the closed disk |s-s0| ≤ ε does not contain any other point of S. Then, by the maximum modulus principle, the maximum value of |g(s)| on the boundary of the disk is attained at some point on the boundary.

Let z be the point on the boundary where the maximum value of |g(s)| is attained. Then, |g(z)| is nonzero, since g(s) = 0 for all s in S except for s0. But this contradicts the fact that g(s) = h(s)/(s-s0), which implies that g(s) approaches zero as s approaches s0.

Therefore, S must have a limit point in the region Re(s) > 0. This completes the proof. The function g(s) = f(s)/(s-s0) is analytic in the region Re(s) > 0, and g(s) = h(s) for some analytic function h(s) in the same region. Therefore, h(s) is also analytic in the region Re(s) > 0.

Since h(s) is analytic in the region Re(s) > 0 and h(s) = 0 for all s in S except for s0, we can write h(s) = (s-s0)k(s), where k(s) is an analytic function in the region Re(s) > 0.

Then, g(s) = h(s)/(s-s0) = k(s) is also analytic in the region Re(s) > 0. Since g(s) = 0 for all s in S except for s0, we have k(s) = 0 for all s in S except for s0.

But this contradicts the assumption that S has no limit point in the region Re(s) > 0, since k(s) = 0 for all s in an open disk around s0.

Therefore, S must have a limit point in the region Re(s) > 0. By the uniqueness property of analytic continuation, it follows that ζ(s) = ζ(1-s*) for all s with Re(s) > 0.

Since the function ζ(s) has an analytic continuation to the entire complex plane except for s = 1, and ζ(s) = ζ(1-s*) for all s with Re(s) > 0, it follows that ζ(s) = ζ(1-s*) for all s except s = 1. This completes the proof of the converse direction.

Therefore, we have shown that the super symmetric equation is true if and only if ζ(s) and ζ(1-s*) are equal to each other. If this equation is not true and the difference of the two series is not zero, but ζ(s) - ζ(1-s*) = 0, then it would mean that two different values have been mapped to the same value, violating the uniqueness property of analytic continuation.

The proof of the super symmetric equation is as follows:

Suppose that ζ(s) and ζ(1-s*) are equal to each other. Then we can write the difference of the two series ∑(n=1 to infinity) 1/n^s and ∑(n=1 to infinity) 1/n^(1-s*) as follows:

∑(n=1 to infinity) (1/n^s - 1/n^(1-s*)) = ∑(n=1 to infinity) (1/n^s(1 - n^(1-2s)))

Using the bound |1 - n^(1-2s)| ≤ 2 for n ≥ 1 and s in the critical strip, we can write:

|∑(n=1 to infinity) (1/n^s(1 - n^(1-2s)))| ≤ 2∑(n=1 to infinity) (1/n^σ)

where σ is the real part of s.

The series ∑(n=1 to infinity) (1/n^σ) converges for σ > 1, so by the Cauchy condensation test, the series ∑(n=1 to infinity) (1/n^s(1 - n^(1-2s))) converges for s in the critical strip. Thus, the difference of the two series converges to zero in the critical strip.

Conversely, suppose that the difference of the two series converges to zero in the critical strip. Then we can write:

ζ(s) - ζ(1-s*) = 2(2π)^(-s) sin(πs/2) ∑(n=1 to infinity) (n^(-s) - n^(s-1))

Using the bound |n^(-s) - n^(s-1)| ≤ 2n^(-σ) for n ≥ 1 and s in the critical strip, we can write:

|ζ(s) - ζ(1-s*)| ≤ 4(2π)^(-σ) sin(πσ/2) ∑(n=1 to infinity) (1/n^σ)

which converges for σ > 1, so ζ(s) and ζ(1-s*) are equal to each other in the critical strip.

Finally, suppose that ζ(s) - ζ(1-s*) = 0, but the difference of the two series is not zero. Then we have two different values of the function mapped to the same value, violating the uniqueness property of analytic continuation. Therefore, the super symmetric equation is an important result in the theory of the Riemann zeta function.

Let $s$ and $1-s^*$ be complex numbers such that $\text{Re}(s) \in (0, 1)$. Then, by the definition of the Riemann zeta function, we have

$$\begin{align} \zeta(s) &= \sum_{n=1}^\infty \frac{1}{n^s} \\\\\\ &= \sum_{n=1}^\infty \frac{1}{n^{1-s^*}} \end{align}$$

Subtracting these two equations, we get

$$\zeta(s) - \zeta(1-s^*) = \sum_{n=1}^\infty \left( \frac{1}{n^s} - \frac{1}{n^{1-s^*}} \right)$$

Now, we note that

$$\begin{align} \frac{1}{n^s} - \frac{1}{n^{1-s^*}} &= \frac{n^{1-s^*} - n^s}{n^{s(1-s^*)}} \\\\ &= \frac{n^{s(1-s^*)} - n^s}{n^{s(1-s^*)}} \cdot \frac{1}{n^s} \\\\ &= \frac{n^s - n^{2s} + n^{3s} - \cdots}{n^{s(1-s^*)}} \cdot \frac{1}{n^s} \\\\ &= \frac{1 - n^s + n^{2s} - \cdots}{n^{s(1-s^*)}} \cdot \frac{1}{n^s} \end{align}$$

Since $\text{Re}(s) \in (0, 1)$, we have $n^s \to 0$ as $n \to \infty$. Therefore, the above expression converges to

$$\frac{1}{n^{s(1-s^*)}} \cdot \frac{1}{n^s} = \frac{1}{n^{s+s(1-s^*)}} = \frac{1}{n^{2s-1}}$$

Therefore, we have

$$\zeta(s) - \zeta(1-s^*) = \sum_{n=1}^\infty \frac{1}{n^{2s-1}}$$

This series converges by the comparison test, since $\frac{1}{n^{2s-1}} \le \frac{1}{n^2}$ for all $n \ge 1$.

Now, let $s_1$ and $s_2$ be complex numbers such that $\text{Re}(s_1) \in (0, 1)$, $\text{Re}(s_2) \in (0, 1)$, and $\zeta(s_1) = \zeta(s_2)$. Then, by the above, we have

$$\sum_{n=1}^\infty \frac{1}{n^{2s_1-1}} = \sum_{n=1}^\infty \frac{1}{n^{2s_2-1}}$$

This implies that the difference between the sums of the series converges to zero.

Conversely, let $s$ be a complex number such that $\text{Re}(s) \in (0, 1)$ and the difference between the sums of the series converges to zero. Then, by the above, we have

$$\zeta(s) - \zeta(1-s^*) = \sum_{n=1}^\infty \left( \frac{1}{n^s} - \frac{1}{n^{1-s^*}} \right) = 0$$

This implies that $\zeta(s) = \zeta(1-s^*)$.

Therefore, the supersymmetric equation states that if $\zeta(s) = \zeta(1-s^*)$, then the difference between the sums of the series converges to zero in the critical strip, and the converse is also true.

It's important to note that if this equation is not true and the difference of the two series is not zero, but $\zeta(s) - \zeta(1-s^*) = 0$, then it would mean that two different values have been mapped to the same value, violating the uniqueness property of analytic continuation.

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Assume ζ(s) = ζ(1-s*). We want to show that the series difference ∑(n=1 to ∞) 1/n^s - ∑(n=1 to ∞) 1/n^(1-s*) converges to 0 as s approaches any point on the line Re(s) = 1/2 in the critical strip.

Consider the function f(z) = ζ(z) - ζ(1-z*), which is defined for Re(z) > 1. We can extend f(z) to a larger region using the functional equation of the zeta function:

f(z) = 2^(z) * π^(z - 1) * sin(πz/2) * Γ(1 - z) * ζ(1 - z).

Let's define a closed contour C consisting of a vertical line segment from -R to R, a semicircular arc centered at the origin with radius R, and a vertical line segment from R to -R. We choose R large enough such that C encloses all the positive integer points.

Applying the residue theorem, the integral of f(z) along C is given by:

∮C f(z) dz = 2πi * (sum of residues inside C).

The integral along the circular arc tends to 0 as R goes to infinity since f(z) is bounded. Thus, we have:

∮C f(z) dz = ∫(-R to R) f(x) dx + ∫(circle) f(z) dz.

Now, let's evaluate the integral along the vertical line segments. As R goes to infinity, the integral on the upper segment approaches the sum ∑(n=1 to ∞) 1/n^s, and the integral on the lower segment approaches the sum ∑(n=1 to ∞) 1/n^(1-s*). Therefore, we have:

∮C f(z) dz = ∑(n=1 to ∞) 1/n^s - ∑(n=1 to ∞) 1/n^(1-s*).

Now, let's evaluate the integral along the horizontal line segment from -R to R. By applying the residue theorem, we find that the integral approaches 2πi times the sum of residues at the poles of f(z) inside the contour C.

The poles of f(z) are located at z = 1, z = 2, z = 3, and so on. The residue at z = n is given by:

Res[f(z), z = n] = (1/n^s - 1/n^(1-s*)).

Since the series difference converges to 0 as s approaches any point on the line Re(s) = 1/2, it follows that the sum of residues at the poles approaches 0.

Therefore, as R goes to infinity, we have:

2πi * (sum of residues) = ∑(n=1 to ∞) 1/n^s - ∑(n=1 to ∞) 1/n^(1-s*) + 0.

Hence, we conclude that the series difference ∑(n=1 to ∞) 1/n^s - ∑(n=1 to ∞) 1/n^(1-s*) converges to 0 as s approaches any point on the line Re(s) = 1/2 in the critical strip.

This completes the proof using contour integration.

sse22.pdf

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**Theorem:** If ζ(s) and ζ(1-s*) are equal to each other, then the difference between the sums of the series ∑(n=1 to infinity) 1/n^s and ∑(n=1 to infinity) 1/n^(1-s*) must converge to zero in the critical strip.

**Proof:**

- Let f(z) = ζ(z) - ζ(1-z*). Then f(z) is defined for Re(z) > 1.
- We can extend f(z) to a larger region, which is the entire critical strip. This can be done using the functional equation of the Riemann zeta function.
- Now, let's consider the series ∑(n=1 to infinity) 1/n^s and ∑(n=1 to infinity) 1/n^(1-s*). These series converge for Re(s) > 1.
- We can then show that the difference between these two series is equal to f(s).
- Since f(z) is continuous on the critical strip, it follows that the difference between the two series must converge to 0 as s approaches any point on the line Re(s) = 1/2 in the critical strip.

**Corollary:** The converse of the theorem is also true. That is, if the difference between the sums of the series ∑(n=1 to infinity) 1/n^s and ∑(n=1 to infinity) 1/n^(1-s*) converges to 0 as s approaches any point on the line Re(s) = 1/2 in the critical strip, then ζ(s) and ζ(1-s*) must be equal to each other.

**Proof:**

Suppose that the difference between the sums of the series ∑(n=1 to infinity) 1/n^s and ∑(n=1 to infinity) 1/n^(1-s*) converges to 0 as s approaches any point on the line Re(s) = 1/2 in the critical strip. Then, by the theorem, f(s) must converge to 0 as s approaches any point on the line Re(s) = 1/2 in the critical strip.

Since f(z) is continuous on the critical strip, it follows that f(z) must be equal to 0 for all z on the line Re(s) = 1/2 in the critical strip.

But f(z) = ζ(z) - ζ(1-z*). Therefore, it follows that ζ(z) = ζ(1-z*) for all z on the line Re(s) = 1/2 in the critical strip.

This completes the proof.

sse24.pdf

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sse25.pdf

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Proofs below have been verified and may contain errors and cannot prove SSE.

sse221 .pdf

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sse23.pdf

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