Coś z pi

[Eariu52]

\(\displaystyle{ \sum_{n=2}^{ \infty } \frac{n+\left( -1\right) ^{n} }{n ^{4}-n ^{2} } = \frac{ \pi ^{2}-6 }{12} }\)

[stefcio2]

\(\displaystyle{ \sum_{n=2}^{\infty}\frac{n+\left(-1\right)^{n}}{n^{4}-n^{2}}=\sum_{n=2}^{\infty}\frac{1}{n^{3}-n}+\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}-n^{2}}}\)

\(\displaystyle{ \sum_{n=2}^{\infty}\frac{1}{n^{3}-n}=\frac{1}{2}\sum_{n=2}^{\infty}\left(\frac{1}{n-1}-\frac{1}{n}+\frac{1}{n+1}-\frac{1}{n}\right)=\frac{1}{2}\cdot\frac{1}{2}=\frac{1}{4}}\)

\(\displaystyle{ \sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}-n^{2}}=\sum_{n=2}^{\infty}\left(\left(-1\right)^{n}\left(\frac{1}{2\left(n-1\right)}-\frac{1}{2\left(n+2\right)}-\frac{1}{n^{2}}\right)\right)=\frac{1}{2}-\frac{1}{4}-1+\left(1-\frac{1}{2^{2}}+\frac{1}{3^{2}}-\frac{1}{4^{2}}+...\right)=\frac{1}{2}-\frac{1}{4}-1+\frac{\pi^{2}}{12}=\frac{\pi^{2}-9}{12}}\)

\(\displaystyle{ \sum_{n=2}^{\infty}\frac{1}{n^{3}-n}+\sum_{n=2}^{\infty}\frac{\left(-1\right)^{n}}{n^{4}-n^{2}}=\frac{1}{4}+\frac{\pi^{2}-9}{12}=\frac{\pi^{2}-6}{12}}\)

[Eariu52]

\(\displaystyle{ \frac{8 \pi }{3}= \sum_{n=0}^{ \infty } \frac{1}{\left( n+ \frac{1}{4} \right) \left( n+ \frac{1}{2} \right)\left( n+1\right) } }\)

[Eariu52]

\(\displaystyle{ \frac{9 \pi }{2} = \sum_{n=0}^{ \infty } \frac{\left( -1\right) ^{n} }{\left( n + \frac{1}{6} \right) \left( n + \frac{3}{6} \right) \left( n + \frac{5}{6} \right) } }\)

[Eariu52]

\(\displaystyle{ 6 \sqrt{3} \pi = \sum_{n=0}^{ \infty } \frac{1}{\left( n+ \frac{1}{6} \right)\left( n+ \frac{2}{6} \right) \left( n+ \frac{4}{6} \right) \left( n+ \frac{5}{6} \right) } }\)

[Eariu52]

\(\displaystyle{ \frac{ \sqrt{3} \pi }{2} = \sum_{n=0}^{ \infty } \frac{\left( -1\right) ^{n} }{\left( n+ \frac{1}{3} \right)\left( n+1\right) } }\)