This proof was released this year:


Where is the mistake? I just found it and was wondering how obviously wrong it is.


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  • 4
    $\begingroup$ Already the very first sentence of the introduction has two mistakes in "every nontirivial zeros" $\endgroup$ – Hagen von Eitzen Nov 23 '15 at 10:16
  • $\begingroup$ @HagenvonEitzen Haha, I mean mathematical mistakes. $\endgroup$ – user85798 Nov 23 '15 at 10:31
  • 1
    $\begingroup$ Nine pages certainly does not induce confidence 😉 $\endgroup$ – Markus Shepherd Nov 23 '15 at 20:19

The proof of lemma 2.1 (the first thing the author proves at all) is false (though I didn't check if the claim of the lemma is possibly correct for other reasons):

Let $R_n(s):= \frac{1+(-1)^n}{n^{s}}$. This $R$ has sufficiently similar properties as the series remainder term, at least as far as they were used in the alleged proof: First of all, $$\tag1\lim_{n\to\infty}R_n(s)=0.$$ Next, $$R_n(s)-R_{n-1}(s)=\frac{1+(-1)^n}{n^{s}}-\frac{1-(-1)^n}{(n-1)^{s}}=\begin{cases}-\frac{2}{(n-1)^s}&n\text{ odd}\\\hphantom{-}\frac2{n^s}&n\text{ even}\end{cases}$$ and $$R_n(s)-R_{n+1}(s)=\frac{1+(-1)^n}{n^{s}}-\frac{1-(-1)^n}{(n+1)^{s}}=\begin{cases}-\frac{2}{(n+1)^s}&n\text{ odd}\\\hphantom{-}\frac2{n^s}&n\text{ even}\end{cases}$$
so that $$\frac{R_n(s)-R_{n-1}(s)}{R_n(s)-R_{n+1}(s)}=\begin{cases}\left(1+\frac2{n-1}\right)^s&n\text{ odd}\\1&n\text{ even} \end{cases}$$ and $$\tag2 \lim_{n\to\infty}\frac{R_n(s)-R_{n-1}(s)}{R_n(s)-R_{n+1}(s)}=1.$$ It appears that the author thinks that $(1)$ and $(2)$ (or maybe even $(2)$ alone) imply $$ \lim_{n\to\infty}\frac{-R_{n-1}(s)}{R_n(s)}= \lim_{n\to\infty}\frac{-R_{n+1}(s)}{R_n(s)}=1.$$ However, with $R$ as defined by me, these limits do not even exist.

  • $\begingroup$ Ha, thanks. What an odd thing to say. $\endgroup$ – user85798 Nov 24 '15 at 3:38

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