e = 5/2 + sigma
: 29 cze 2024, o 05:50
\(\displaystyle{ e = \frac{5}{2} + \sum_{n=0}^{ \infty } \frac{1}{n!\left( n+2\right) \left( n+4\right) } }\)
$$\frac{1}{n!(n+2)(n+4)}=\frac{1}{(n+2)!}-\frac{3}{(n+3)!}+\frac{3}{(n+4)!},$$
więc po steleskopowaniu drugich i trzecich składników wychodzi
$$\sum_{n=0}^\infty \frac{1}{n!(n+2)(n+4)} = \sum_{n=0}^\infty \frac{1}{(n+2)!} - \frac{3}{3!} = \left(e-\frac{1}{0!}-\frac{1}{1!}\right)-\frac 12 = e-\frac 52$$