I was wondering if anyone is aware of this proof about the Riemann Hypothesis. https://arxiv.org/pdf/1004.4143.pdf
What I like about it is his analytic approach using the behaviour of general series at singular points. Since I haven't heard anything about it, I presume there must be some flaw somewhere, but I can not find it. To me it seems legit. In particular I want to point out his main result. I added short variations of the proofs which should be sufficient to capture the entire essence. The paper with that regard is somewhat cumbersome.
He resums the $\zeta$-function, compares it with its integral representation and obtains an expression valid for $\sigma>0$ ($s=\sigma+it$)
$$(s-1)\zeta(s) = \sum_{n=1}^\infty \frac{S_n(s)}{n+1} \tag{0}$$
where $S_n(s)=-\frac{1}{n} \sum_{k=1}^n \binom{n}{k} (-1)^k \, k^{1-s}$.
Then he defines a function $\zeta(s,x)$ valid for $|x|<1$ by
$$(s-1)\zeta(s,x) = \sum_{n=1}^\infty \frac{S_n(s)}{n+1} \, x^n$$
which converges to (0) for $\sigma>0$ as $x\rightarrow 1$. He uses a result for the asymptotic behaviour of $S_n(s)$ as $n$ gets large $$\frac{S_n(s)}{n+1} \sim \frac{1}{n(n+1)(\log n)^{1-s} \Gamma(s)} \left( 1 + {\cal O}\left(\frac{1}{\log n}\right) \right)\tag{as.1}$$
which probably inspired him to look for functions $\phi_s(x)=\sum_{n=0}^\infty \phi_n(s) \, x^n$ whose coefficients $\phi_n(s)$ have the same asymptotic behaviour with respect to $n$. Such a function is $$\phi_s(x)=(1-x)\left(\frac{\log(1-x)}{-x}\right)^{s}$$
whose coefficients obey $$\phi_n(s) \sim \frac{-s}{n^2 (\log n)^{1-s}} \left( 1 + {\cal O}\left(\frac{1}{\log n}\right) \right)\tag{as.2}$$
as $n$ gets large.
Since $\phi_s'(x)$ diverges to $\infty$ as $x\rightarrow 1$ it follows that $\zeta'(s,x)$ diverges to $\infty$ as $x\rightarrow 1$. This allows him to establish the result (after a lot of details) $$\lim_{x\rightarrow 1} \frac{(s-1)\zeta'(s,x)}{\phi_s'(x)} = \lim_{n\rightarrow \infty} \frac{\frac{nS_n(s)}{n+1}}{n\phi_n(s)} = \frac{-1}{\Gamma(s+1)} \, . \tag{1}$$
In a similar but different way as in the pre-print this can be easily seen as follows.
proof: Define $a_n\equiv\frac{nS_n(s)}{n+1}$ and $b_n\equiv n\phi_n(s)$, so that $\lim_{n\rightarrow \infty}\frac{a_n}{b_n} = \frac{-1}{\Gamma(s+1)} \equiv c$. This is equivalent to saying $\frac{a_n}{b_n}=c+\epsilon_n$ for some complex sequence $\epsilon_n$ that vanishes as $n\rightarrow \infty$. From the asymptotics (as.1) and (as.2) it is seen that for large $n$ $$|\epsilon_n| = {\cal O}\left(1/\log n\right) \, .$$ To establish the result, $a_n = cb_n + \epsilon_n b_n$ is used to construct $$\sum_{n=1}^\infty a_n x^n = c \sum_{n=1}^\infty b_n x^n + \sum_{n=1}^\infty \epsilon_n b_n x^n$$ from which it then follows if we can show that $\sum_{n=1}^\infty |\epsilon_n b_n|$ converges. But from the asymptotics for $|\epsilon_n|$ and $|b_n|$ this sum can be bounded $$\sum_{n=2}^\infty |\epsilon_n b_n| \leq C|s| \sum_{n=2}^\infty \frac{1}{n(\log n)^{2-\sigma}}$$ for a suitable constant $C<\infty$. The convergence of the right sum follows by the integral test for $0<\sigma<1$.
From this it can be deduced: If $s$ is a non-trivial zero, then $$\lim_{x\rightarrow 1} \frac{(s-1)\zeta(s,x)}{\phi_s(x)} = \frac{-1}{\Gamma(s+1)} \, . \tag{2}$$
proof: Define $\delta(x)\equiv \frac{(s-1)\zeta'(s,x)}{\phi_s'(x)} + \frac{1}{\Gamma(s+1)}$, so that the previous result is compactly written as $\lim_{x\rightarrow 1} \delta(x) = 0$. Let $s$ be a non-trivial zero and integrate (the previous proof showed that $\zeta'(s,x)$ has the same singularity at $x=1$ as $\phi_s'(x)$ which is manifestly integrable in a neighbourhood $\leq 1$) $$\int_x^1 \delta(t) \phi_s'(t) \, {\rm d}t = -(s-1)\zeta(s,x) - \frac{\phi_s(x)}{\Gamma(s+1)}$$ or $$-\frac{\int_x^1 \delta(t) \phi_s'(t) \, {\rm d}t}{\phi_s(x)} = \frac{(s-1)\zeta(s,x)}{\phi_s(x)} + \frac{1}{\Gamma(s+1)} \, .$$ It suffices to show that the LHS vanishes absolutely as $x\rightarrow 1$, for this $$\left|\frac{\int_x^1 \delta(t) \phi_s'(t) \, {\rm d}t}{\phi_s(x)}\right| \leq \frac{\int_x^1 |\delta(t)| \, |\phi_s'(t)| \, {\rm d}t}{|\phi_s(x)|} \, .$$ The nominator ($f(x)$) and denominator ($g(x)$) on the RHS are differentiable on the open interval $(x,1)$ and finite at $x<1$, so Rolle's theorem (for the function $h(t)=f(t) - \frac{f(x)}{g(x)} \, g(t)$) or equivalently L'Hospital can be applied. The RHS becomes in the limit $x\rightarrow 1$ $$\lim_{x\rightarrow 1} |\delta(x)| \, \frac{|\phi_s'(x)|}{-|\phi_s(x)|'} = 0 \cdot 1 = 0 \, .$$
In the final step he considers the ratio of (2) for the zero $s$ and $1-s$
$$ \lim_{x\rightarrow 1} \left|\frac{(s-1)\zeta(s,x)}{\phi_s(x)}\frac{\phi_{1-s}(x)}{(-s)\zeta(1-s,x)}\right| = \left|\frac{\Gamma(2-s)}{\Gamma(s+1)}\right| \tag{3}$$
or
$$ \lim_{x\rightarrow 1} \left|\frac{\phi_{1-s}(x)}{\phi_s(x)}\right| = \lim_{x\rightarrow 1} \left( \frac{\log(1-x)}{-x} \right)^{1-2\sigma} = \left|\frac{\zeta(1-s)\Gamma(1-s)}{\zeta(s)\Gamma(s)}\right| $$
which is an equation only possible when $\sigma =1/2$ in which case the LHS is 1, which is indeed true for any real $t$ on the RHS when using the functional equation for the $\zeta$-function
$$\left| \frac{\zeta(1-s)\Gamma(1-s)}{\zeta(s)\Gamma(s)} \right| = \left| \frac{(2\pi)^{1-s}}{2\sin(\pi s/2)\Gamma(s)} \right| = 1 $$
and $\left|\Gamma(1/2+it)\right|^2 = \frac{\pi}{\cosh(\pi t)}$.
Wrong... As most probably have expected, the proof is fundamentally flawed. I've been thinking about this for some time now and am somewhat embarassed to say that the solution is so trivial. It would have saved me lots of time, if somebody here would have pointed it out earlier, since it could have been seen from my given exposition, but here we go...
The problem with the proof lies within its formulation. It utterly overcomplicates matter where not necessary. This distracts from the relevant point and superficially makes it look as if it were correct, just like shenanigans try to distract their prey. While I have already simplified the steps significantly (I believe), it already more or less starts with Equation (1). It is not even wrong, but should have been more properly written as $$(s-1)\zeta'(s,x) = -\frac{\phi_s'(x)}{\Gamma(s+1)} + {\cal O}(1)$$ for $x\rightarrow 1$, in order to keep track of error-terms. This shows that both functions share the same logarithmic singularity at $x=1$.
Now when integrating this Equation from $x$ to $1$, it is not yet assumed that $s$ is a zero. This gives rise to the identity $$(s-1)\zeta(s,x)=(s-1)\zeta(s) - \frac{\phi_s(x)}{\Gamma(s+1)} + {\cal O}(1-x)$$ as $x\rightarrow 1$, which is just the expansion of $(s-1)\zeta(s,x)$ about the singular point $x=1$.
Now if $s=s_0$ is a zero with $0<\Re(s_0)<1$, then the constant term of the expansion vanishes identically, so that Equation (3) becomes $$\lim_{x\rightarrow 1} \left|\frac{(s_0-1)\zeta(s_0,x)}{(-s_0)\zeta(1-s_0,x)}\right| = \lim_{x\rightarrow 1} \left| \frac{\phi_{s_0}(x)}{\phi_{1-s_0}(x)} \frac{\Gamma(2-s_0)}{\Gamma(s_0+1)}\right| \\ \stackrel{!}{=} \lim_{s\rightarrow s_0} \left| \frac{(s-1)\zeta(s,1)}{(-s)\zeta(1-s,1)} \right| = \lim_{s\rightarrow s_0} \left| \frac{(s-1)\zeta(s)}{(-s)\zeta(1-s)} \right| \, .$$
This is where the flaw becomes apparent, because the interchange of limits is generally wrong and would have to be crucially justified. Evidently it would be the main part of the proof!
I should also mention, that Equation (0) - even though he might have found it independently - can actually be traced back to a version from H. Hasse (1930).