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 $r=\lim _{ n\rightarrow \infty }{ \left| \frac { a_{ { n } } }{ a_{ { n+1 } } } \right| } =e$

-
Thanks,i used the first formula on a) and found out that it is unlimited ($\infty$) so it does not have a convergence radius. With the same formula i found it difficult to solve b) – Devid Jan 27 '13 at 18:53
For $b$, use the root test which implies directly $|z|<1$. – Mhenni Benghorbal Jan 27 '13 at 19:04

For a), the series diverges $\forall \, z \ne 0$ because $\lim_{n \rightarrow \infty} (2 + \sqrt{n}) = \infty$.
$$\lim_{n \rightarrow \infty} \left ( 1-\frac{1}{n} \right )^n = \frac{1}{e}$$
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$. - Thanks for the super fast answer, so i need to use geometric series to show b) right ? I will try that now and solve b) – Devid Jan 27 '13 at 18:54 @Devid: you don't need to "use" the geometric series; all you need to do is recognize that a series of the form$a(1+z+z^2+\ldots$, i.e. one with constant coefficients, is a geometric series, and you know that its radius of convergence is$1$. – Ron Gordon Jan 27 '13 at 18:56 If you wish, can use Root Test mechanically on b), the$n$-th root of$a_n$is$1-1/n$, which has limit$1$. The more concrete answer of rigordonna is better. – André Nicolas Jan 27 '13 at 19:20 @rlgordonma i see now, but i thought the radius of convergence of the geometric series is$\frac { 1 }{ 1-q } $. Also could you explain how you come to$\frac{1}{e}$– Devid Jan 27 '13 at 19:37 @Devid: Zo, I think you are quoting the actual sum of a geometric series (I may be wrong, I don't know what$q$is); that sum is only valid for$|z|<1$. The reason is that a series representation is valid up to the pole closest to the origin (or from whereever you are basing your series), and the closest pole is at$z=1$. The limit comes from the definition of$e^{z}\$. – Ron Gordon Jan 27 '13 at 20:11