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$$\sum_{n=1}^{\infty}\frac{1}{1+n^2}\approx1.07169$$

I have viewed the proof of the irrationality of $e$ and $\frac{\pi^2}{6}$ but no idea with this one.

yuanming luo
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    If $\pi\coth(\pi)$ is irrational, then that sum is also irrational, since we have that: $$\sum_{n=1}^{\infty} \frac{1}{a^2+n^2}=\frac{a\pi\coth(a\pi)-1}{2a^2}$$ – projectilemotion Sep 16 '19 at 23:50
  • Thanks, That answers the question – yuanming luo Sep 17 '19 at 00:00
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    Well, I don't think it really answers the question because proving that $\pi\coth(\pi)$ is irrational seems very non-trivial (although intuitively, it may seem like it). My first comment was just to show that we could reduce the problem to a (possibly) simpler one. – projectilemotion Sep 17 '19 at 00:22
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    Usually , irrationality proofs are extremely hard (this does not necessarily mean that this is also the case here). Examples of numbers for which is unknown whether they are rational include $\gamma$ , $e+\pi$ and $e \cdot \pi$ – Peter Sep 17 '19 at 07:51
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    It is known that $\pi$ and $\coth(\pi)$ are irrational. $\coth(\pi)$ is irrational, since we have that: $$\coth(\pi)=1+\frac{2}{e^{2\pi}-1}$$ And $e^{2\pi}$ is transcendental (hence irrational), since $e^{2\pi}=(-1)^{-2i}$, so by the Gelfond–Schneider theorem, the result follows. However, from this one cannot conclude that their product (so in this case $\pi\coth(\pi)$) is irrational (do you know why?). If I had to guess, I would guess that this problem is unsolved. – projectilemotion Sep 17 '19 at 10:45
  • So, the product of two transcendental numbers can be rational? @projectilemotion – yuanming luo Sep 17 '19 at 21:10
  • Yes, consider for instance $a=e$ and $b=1/e$ (reciprocal of a transcendental number is transcendental), but $ab=1$. More generally, you could replace $e$ by any transcendental number. – projectilemotion Sep 17 '19 at 21:41

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This value is demonstrably irrational and transcendental. Nesterenko showed $\pi$ and $e^{\sqrt{a}\pi}$ are algebraically independent over $\mathbb{Q}$ for $a \in \mathbb{N}$ which implies $$\pi\coth\left(\pi\sqrt{a}\right)=\pi\frac{e^{2\pi\sqrt{a}}+1}{e^{2\pi\sqrt{a}}-1}$$ is transcendental which implies

$$\sum_{n=1}^\infty\frac{1}{n^2+a}=\frac{\sqrt{a}\pi\coth(\sqrt{a}\pi)-1}{2a}$$ is transcendental for $a\in\mathbb{N}$. In particular, this means that the value in the title where $a=1$ is irrational.

Mason
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