Задача 64930 1 курс- мат.анализ...
Условие
Решение
[m]a_{n+1}=\frac{9^{n+1}\cdot (n+1)!\cdot (2(n+1)+1)!\cdot (2(n+1))!!}{7^{n+1}\cdot((( n+1)+1)!)^2\cdot (4(n+1)-3)!!}[/m]
[m]R=lim_{n → ∞ }\frac{a_{n}}{a_{n+1}}=lim_{n → ∞ }\frac{\frac{9^{n}\cdot n!\cdot (2n+1)!\cdot (2n)!!}{7^{n}\cdot(( n+1)!)^2\cdot (4n-3)!!}}{\frac{9^{n+1}\cdot (n+1)!\cdot (2(n+1)+1)!\cdot (2(n+1))!!}{7^{n+1}\cdot((( n+1)+1)!)^2\cdot (4(n+1)-3)!!}}=[/m]
[m]=lim_{n → ∞ }\frac{9^{n}\cdot n!\cdot (2n+1)!\cdot (2n)!!}{7^{n}\cdot(( n+1)!)^2\cdot (4n-3)!!}\cdot \frac{7^{n+1}\cdot((( n+1)+1)!)^2\cdot (4(n+1)-3)!!}{9^{n+1}\cdot (n+1)!\cdot (2(n+1)+1)!\cdot (2(n+1))!!}=[/m]
[m]=lim_{n → ∞ }\frac{7}{9}\frac{n!}{(n+1)!}\cdot \frac{(2n+1)!}{(2n+3)!}\cdot\frac{(2n)!!}{(2n+2)!!}\cdot \frac{(( n+2)!)^2}{( n+1)!)^2}\cdot \frac{ (4n+1)!!}{ (4n-3)!!}=[/m]
[m](n+1)!=n!\cdot (n+1)[/m]
[m](2n+3)!=(2n+1)!\cdot (2n+2)\cdot (2n+3)[/m]
[m](2n)!!=2\cdot 4\cdot 6\cdot ...\cdot (2n-2)\cdot (2n)[/m]
[m](2n+2)!!=2\cdot 4\cdot 6\cdot ...\cdot (2n-2)\cdot (2n)\cdot (2n+2)[/m]
[m](n+2)!=((n+1)!)\cdot (n+2)
[m]((n+2)!)^2=((n+1)!)^2\cdot (n+2)^2[/m]
[m](4n-3)!!=1\cdot 5\cdot 9\cdot ...\cdot (4n-7) \cdot (4n-3)[/m]
[m](4n+1)!!=1\cdot 5\cdot 9\cdot ...\cdot (4n-7) \cdot (4n-3)\cdot (4n+1)[/m]
Тогда
[m]=lim_{n → ∞ }\frac{7}{9}\frac{1}{(n+1)}\cdot \frac{1}{(2т+2)(2n+3)}\cdot\frac{1}{(2n+2)}\cdot \frac{(n+2)^2}{1}\cdot \frac{4n+1 }{1}=0[/m]
Ряд сходится в одной точке