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

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  • $\begingroup$ 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) $\endgroup$
    – Devid
    Jan 27, 2013 at 18:53
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    $\begingroup$ For $b$, use the root test which implies directly $|z|<1$. $\endgroup$ Jan 27, 2013 at 19:04

1 Answer 1

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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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  • $\begingroup$ 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) $\endgroup$
    – Devid
    Jan 27, 2013 at 18:54
  • $\begingroup$ @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 $1$. $\endgroup$
    – Ron Gordon
    Jan 27, 2013 at 18:56
  • $\begingroup$ 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. $\endgroup$ Jan 27, 2013 at 19:20
  • $\begingroup$ @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 $\frac{1}{e}$ $\endgroup$
    – Devid
    Jan 27, 2013 at 19:37
  • $\begingroup$ @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 $e^{z}$. $\endgroup$
    – Ron Gordon
    Jan 27, 2013 at 20:11

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