Name: Anonymous 2009-07-13 20:02
I'm trying to find the radius of convergence of \sum_{n=0}^{\infty}{ \frac{1}{n} \cdot e^{n\cdot(-1)^n}\cdot x^n } but I'm apparently too stupid.
The root test delivers: \limsup_{n\rightarrow\infty}\sqrt[n]{\left|\frac{1}{n} \cdot e^{n\cdot(-1)^n}\cdot x^n\right|} = \limsup_{n\rightarrow\infty}\frac{1}{\sqrt[n]{n}} \cdot e^{(-1)^n} \cdot \left|x\right| = e \cdot \left|x\right|
So I could choose \left|x\right| < \frac{1}{e}.
On the other hand I could try the ratio test, which gives: \lim_{n \rightarrow \infty} \left| \frac{n}{n+1} \cdot \frac{e^{(n+1)\cdot(-1)^{n+1}}}{e^{n\cdot(-1)^n}} \cdot \frac{x^{n+1}}{x^n} \right| = \lim_{n \rightarrow \infty} \left| \frac{e^{-n\cdot(-1)^n} \cdot e^{(-1)^{n+1}}}{e^{n\cdot(-1)^n}} \cdot x \right| = \lim_{n \rightarrow \infty} \left| e^{(2n+1)\cdot(-1)^{n+1}}\cdot x\right|
Which diverges, no matter how small I choose x.
Which test do I believe? Where did I go wrong?
The root test delivers: \limsup_{n\rightarrow\infty}\sqrt[n]{\left|\frac{1}{n} \cdot e^{n\cdot(-1)^n}\cdot x^n\right|} = \limsup_{n\rightarrow\infty}\frac{1}{\sqrt[n]{n}} \cdot e^{(-1)^n} \cdot \left|x\right| = e \cdot \left|x\right|
So I could choose \left|x\right| < \frac{1}{e}.
On the other hand I could try the ratio test, which gives: \lim_{n \rightarrow \infty} \left| \frac{n}{n+1} \cdot \frac{e^{(n+1)\cdot(-1)^{n+1}}}{e^{n\cdot(-1)^n}} \cdot \frac{x^{n+1}}{x^n} \right| = \lim_{n \rightarrow \infty} \left| \frac{e^{-n\cdot(-1)^n} \cdot e^{(-1)^{n+1}}}{e^{n\cdot(-1)^n}} \cdot x \right| = \lim_{n \rightarrow \infty} \left| e^{(2n+1)\cdot(-1)^{n+1}}\cdot x\right|
Which diverges, no matter how small I choose x.
Which test do I believe? Where did I go wrong?