Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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
2  
For $b$, use the root test which implies directly $|z|<1$. –  Mhenni Benghorbal Jan 27 '13 at 19:04
add comment

1 Answer

up vote 1 down vote accepted

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 $\forall \, |z| < e$.

share|improve this answer
    
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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.