\(\displaystyle{ \frac{1}{\left( n+1\right)^3\left( n+2\right)^3 } = \red{\frac{6}{n+1}- \frac{6}{n+2}}+\blue{ \frac{1}{\left( n+1\right)^3 }-\frac{1}{\left( n+2\right)^3 } }\green{- \frac{3}{\left( n+1\right)^2 } }-\green{ \frac{3}{\left( n+2\right)^2 } } }\)
\(\displaystyle{ \sum_{n=0}^{N} \frac{1}{\left( n+1\right)^3\left( n+2\right)^3 } = \sum_{n=0}^{N} \left( \red{\frac{6}{n+1}- \frac{6}{n+2}}\right) +\sum_{n=0}^{N}\left( \blue{ \frac{1}{\left( n+1\right)^3 }-\frac{1}{\left( n+2\right)^3 } }\right) - \sum_{n=0}^{N}\green{ \frac{3}{\left( n+1\right)^2 } }- \sum_{n=0}^{N}\green{ \frac{3}{\left( n+2\right)^2 } }}\)
\(\displaystyle{ \sum_{n=0}^{N} \frac{1}{\left( n+1\right)^3\left( n+2\right)^3 } = \red{ \frac{6N+6}{N+2} } + \blue{ \frac{N^3+6N^2+12N+7}{\left( N+2\right)^2 } }- \sum_{n=0}^{N}\green{ \frac{3}{\left( n+1\right)^2 } }- \sum_{n=0}^{N}\green{ \frac{3}{\left( n+2\right)^2 } }}\)
\(\displaystyle{ \sum_{n=0}^{N} \frac{1}{\left( n+1\right)^3\left( n+2\right)^3 } = \red{ \frac{6N+6}{N+2} } + \blue{ \frac{N^3+6N^2+12N+7}{\left( N+2\right)^2 } }- 3\left( \sum_{n=1}^{N+1} \frac{1}{n^2}+ \sum_{n=2}^{N+2} \frac{1}{n^2} \right) }\)
\(\displaystyle{ \sum_{n=0}^{N} \frac{1}{\left( n+1\right)^3\left( n+2\right)^3 } = \red{ \frac{6N+6}{N+2} } + \blue{ \frac{N^3+6N^2+12N+7}{\left( N+2\right)^2 } }- 3\left( \sum_{n=1}^{N+1} \frac{1}{n^2}+ \sum_{n=1}^{N+2} \frac{1}{n^2} -1\right) }\)
Kładąc
\(\displaystyle{ N\to \infty }\) mamy:
\(\displaystyle{ \sum_{n=0}^{ \infty } \frac{1}{\left( n+1\right)^3\left( n+2\right)^3 } = \red{6 } + \blue{ 1}- 3\left( 2 \cdot \frac{ \pi ^2}{6} -1\right)=10-\pi^2 }\)
