$$\sum\limits_{n = 1}^\infty {\frac{{{H_{\left[ {\frac{n}{3}} \right]}}}}{{{n^2}}}{{\left( { - 1} \right)}^{n - 1}}} =$$
$$\frac{1}{12}\log^3(3)-\frac{\pi^2}{36}\log\left(\frac{256}{243}\right)-\frac{7}{24}\zeta(3)+\frac{1}{2}\log(3)\operatorname{Li_2}\left(\frac{1}{6}\left(3-i\sqrt{3}\right)\ri...
If $p$ and $q$ are primes larger than $2$, then $pq+1$ is even.
I am trying to prove the following: If $p$ and $q$ are prime numbers larger than $2$, then $pq + 1 $ is never prime. Any ideas?
First, let $-\log(\sin(\theta))=y$ that yields
$$\underbrace{\int_0^{\infty} \log(y) \log(1-e^{-2 y}) \ dy}_{\displaystyle \pi^2 \log(A)-\frac{1}{12}\pi^2 \log(\pi)}-\frac{1}{2}\underbrace{\int_0^{\infty} \log(\pi^2+y^2) \log(1-e^{-2 y}) \ dy}_{\displaystyle\sum_{k=1}^{\infty} \frac{\operatornam...
We are familiar with the classic sum for Euler's constant $\gamma$:
$$ \gamma=\lim_{n\to \infty}\left(\sum_{k=1}^{n}\frac{1}{k}-\ln(n)\right) .$$
But, how is this one derived?:
$$ \gamma=\lim_{n\to \infty}\left(\frac12\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k(2k)!}n^{2k}-\ln(n)\right) .$$
I thou...
