query: Fault in proof of $\zeta(2) = \frac{\pi^2}{6}$
positive_passages.docid: 2302
positive_passages.text: Different methods to compute $\sum\limits_{k=1}^\infty \frac{1}{k^2}$ (Basel problem)
negative_passages.docid: 36503
negative_passages.text: Nicomachus theorem proof - what did I do wrong?
negative_passages.docid: 5679
negative_passages.text: Why is $2\pi i \neq 0?$
negative_passages.docid: 122932
negative_passages.text: When $(\mathbb{Q}(\zeta_n) : \mathbb{Q}) = 2$?
negative_passages.docid: 120743
negative_passages.text: For Riemann sums involving square roots, why do we let $c_{i} = \frac{i^{2}}{ n^{2}}$?
negative_passages.docid: 137640
negative_passages.text: Proof that $\sum_{1}^{\infty} \frac{1}{n^2} <2$
negative_passages.docid: 48086
negative_passages.text: Why is this series of square root of twos equal $\pi$?
negative_passages.docid: 15210
negative_passages.text: Group of order 6 proof
negative_passages.docid: 136855
negative_passages.text: Proof that $\mathbb{Z}\left[\frac{1 + \sqrt{-19}}{2}\right]$ is a PID
negative_passages.docid: 123056
negative_passages.text: Help in understanding the proof of Mersenne Prime
negative_passages.docid: 24202
negative_passages.text: The value of the trilogarithm at $\frac{1}{2}$
negative_passages.docid: 12112
negative_passages.text: What is known about the pattern for $\zeta(2n+1)$?
negative_passages.docid: 157946
negative_passages.text: Elementary derivation of certian identites related to the Riemannian Zeta function and the Euler-Mascheroni Constant
negative_passages.docid: 131739
negative_passages.text: Is my proof of the additivity property of Riemann integral correct?
negative_passages.docid: 98305
negative_passages.text: How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$
negative_passages.docid: 6436
negative_passages.text: Proving an alternating Euler sum: $\sum_{k=1}^{\infty} \frac{(-1)^{k+1} H_k}{k} = \frac{1}{2} \zeta(2) - \frac{1}{2} \log^2 2$
negative_passages.docid: 67271
negative_passages.text: $\sum_{n=-\infty}^{\infty}\frac{1}{(u +n)^2}=\frac{\pi^2}{(\sin \pi u)^2}$