linguistic:paraphrasing, linguistic:translation
Pi

∑ n = 1 ∞ 1 n 2 = 1 1 2 + 1 2 2 + 1 3 2 + 1 4 2 + ⋯ = π 2 6 {\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {1}{1^{2}}}+{\frac {1}{2^{2}}}+{\frac {1}{3^{2}}}+{\frac {1}{4^{2}}}+\cdots ={\frac {\pi ^{2}}{6}}}

∑ n = 1 ∞ (− 1) n + 1 (4) 2 n − 1 = 4 1 − 4 3 + 4 5 − 4 7 + 4 9 − 4 11 + 4 13 − ⋯ = π {\displaystyle \sum _{n=1}^{\infty }{\frac {(-1)^{n+1}(4)}{2n-1}}={\frac {4}{1}}-{\frac {4}{3}}+{\frac {4}{5}}-{\frac {4}{7}}+{\frac {4}{9}}-{\frac {4}{11}}+{\frac {4}{13}}-\cdots =\pi }