Question

Найти область сходимости ряда: $\displaystyle \large _\infty\\\sum\dfrac{2^nx^n}{\sqrt{n+3} } \\_n_=_0$.

Answer 1

Ответ:

$\displaystyle \sum \limits _{n=0}^{\infty }\, \dfrac{2^{n}\cdot x^{n}}{\sqrt{n+3}}\\\\\\\lim\limits_{n \to \infty}\, \frac{|u_{n+1}|}{|\, u_{n}\, |}=\lim\limits_{n \to \infty}\, \frac{2^{n+1}\cdot |x|^{n+1}}{\sqrt{n+4}}\cdot \frac{\sqrt{n+3}}{2^{n}\cdot |\, x\, |^{n}}=2\cdot |\, x\, |<1\\\\\\|x|<\frac{1}{2}\ \ \ \Rightarrow \ \ \ -\frac{1}{2}<x<\frac{1}{2}\\\\a)\ \ x=\frac{1}{2}:\ \ \sum \limits _{n=0}^{\infty }\, \dfrac{1}{\sqrt{n+3}}\ \ -\ \ rasxoditsya\ ,\ garmonicheskij$

$b)\ \ x=-\dfrac{1}{2}:\ \ \sum \limits _{n=0}^{\infty }\, \dfrac{(-1)^{n}}{\sqrt{n+3}}\ \ -\ \ yslovno\ sxoditsya\\\\\\Otvet:\ \ x\in \Big[-\dfrac{1}{2}\ ;\ \dfrac{1}{2}\ \Big)\ .$