<s>[INST] Power Series and Radius of Convergence 
Determine for following Power Series in $\mathbb{C}$ the radius of
  Convergence.
a) $\sum _{ n=0 }^{ \infty  }{ (2+\sqrt { n } )^{ n }z^{ n } } $
b) $\sum _{ n=1 }^{ \infty  }{ (1-\frac { 1 }{ n } )^{ n }z^{ n } } $
c) $\sum _{ n=0 }^{ \infty  }{ n!n^{ -n }z^{ n } } $
For the radius of convergence, on has the formula: $ r=\frac { 1
> }{ L } ,\quad wo\quad L:=\lim _{ n\rightarrow \infty  }{ sup\sqrt [ n
> ]{ \left| { a }_{ n } \right|  }  }$
Or the simpler formula:  $r=\lim _{ n\rightarrow \infty  }{ \left|
> \frac { { a }_{ n } }{ { a }_{ n+1 } }  \right| }$


Hear are the solutions:
a) ${ a }_{ n }:=(2+\sqrt { n } )^{ n }, \text{for} \ \ n\in\mathbb{N} $ we have
$\sqrt [ n ]{ \left| { a }_{ n } \right|  } =\sqrt [ n ]{ \left| 2+\sqrt { n }  \right| ^{ n } } =\left| 2+\sqrt { n }  \right| $ and so $\lim _{ n\rightarrow \infty  }sup{ \sqrt [ n ]{ \left| { a }_{ n } \right|  }  } =\infty $ and the Convergence Radius is $r=\frac { 1 }{ \lim _{ n\rightarrow \infty  } sup{ \sqrt [ n ]{ \left| { a }_{ n } \right|  }  } } =0$
b) ${ a }_{ n }:=(1-\frac { 1 }{ n } )^{ n }$ for $n\in \mathbb{N}$ we have $\sqrt [ n ]{ \left| { a }_{ n } \right|  } =\sqrt [ n ]{ \left| 1-\frac { 1 }{ n }  \right| ^{ n } } =1-\frac { 1 }{ n } $ and so $\lim _{ n\rightarrow \infty  }{ sup\sqrt [ n ]{ \left| { a }_{ n } \right| } } =1$ and the Convergence Radius is $r=\frac { 1 }{ \lim _{ n\rightarrow \infty  } sup{ \sqrt [ n ]{ \left| { a }_{ n } \right|  }  } } =1$
c) $a_{n}:=n!n^{-n} \neq0$ for all $n\in\mathbb{N}$ and $\left| \frac { a_{ { n } } }{ a_{ { n }+1 } }  \right| =\frac { n!(n+1)^{ { n+1 } } }{ n^{ n }(n+1)! } =\frac { 1 }{ n+1 } \frac { (n+1)^{ n+1 } }{ n^{ n } } =(\frac { n+1 }{ n } )^{ n }=(1+\frac { 1 }{ n } )^{ n }$ we know that $\lim _{ n\rightarrow \infty  }{ (1+\frac { 1 }{ n } ) } ^{ n }=e$ so the Convergence Radius is $r=\lim _{ n\rightarrow \infty  }{ \left| \frac { a_{ { n } } }{ a_{ { n+1 } } }  \right|  } =e$
 [/INST] For a), the series diverges $\forall \, z \ne 0$ because $\lim_{n \rightarrow \infty} (2 + \sqrt{n}) = \infty$.
For b)
$$\lim_{n \rightarrow \infty} \left ( 1-\frac{1}{n} \right )^n = \frac{1}{e}$$
so by the comparison test with a geometric series, the radius of convergence is $1$, i.e. the series converges only when |z| < 1$.
For c), use Stirling's approximation and the root test to show that the series converges $\forall \, |z| < e$.
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