Suma odwrotności iloczynów
: 21 cze 2025, o 04:40
\(\displaystyle{ \sum_{n=1}^{ \infty } \frac{1}{ \prod_{k=1}^{n} \left( k+a- \frac{1}{k+a} \right)} = \frac{1}{a} , a > 0 }\)
\(\displaystyle{ S _{1}= \sum_{n=1}^{ \infty } \frac{1}{ \prod_{k=1}^{2n-1} \left( k+1- \frac{1}{k+1} \right) } }\)
\(\displaystyle{ S _{2}= \sum_{n=1}^{ \infty } \frac{1}{ \prod_{k=1}^{2n} \left( k+1- \frac{1}{k+1} \right) } }\)
\(\displaystyle{ \frac{S _{2} }{S _{1} }= \frac{e}{2} - 1 }\)