Coś z pi
: 26 wrz 2025, o 04:16
\(\displaystyle{ \sum_{n=2}^{ \infty } \frac{n+\left( -1\right) ^{n} }{n ^{4}-n ^{2} } = \frac{ \pi ^{2}-6 }{12} }\)
\(\displaystyle{ \sum_{n=2}^{\infty}\frac{n+\left(-1\right)^{n}}{n^{4}-n^{2}}=\sum_{n=2}^{\infty}\frac{1}{n^{3}-n}+\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}-n^{2}}}\)
\(\displaystyle{ \sum_{n=2}^{\infty}\frac{1}{n^{3}-n}=\frac{1}{2}\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n+1}-\frac{1}{n}\right)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}}\)
\(\displaystyle{ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}-n^{2}}=\sum_{n=2}^{\infty}\left(\left(-1\right)^{n}\left(\frac{1}{2\left(n-1\right)}-\frac{1}{2\left(n+2\right)}-\frac{1}{n^{2}}\right)\right)=\frac{1}{2}-\frac{1}{4}-1+\left(1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+...\right)=\frac{1}{2}-\frac{1}{4}-1+\frac{\pi^{2}}{12}=\frac{\pi^{2}-9}{12}}\)
\(\displaystyle{ \sum_{n=2}^{\infty}\frac{1}{n^{3}-n}+\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}-n^{2}}=\frac{1}{4}+\frac{\pi^{2}-9}{12}=\frac{\pi^{2}-6}{12}}\)