Is the radius of convergence 1 or 1/2?

[lucad93]

Hello! I have a problem with the following exercise, in which i must calculate the ray of a power serie. This is the power serie: \(\displaystyle \sum_{K=0}^{+\infty}(k+1)z^{k+1}\). I decide to use the ratio test, and so i calculate \(\displaystyle \lim_{k\rightarrow +\infty}\frac{a_{n+1}}{a_{n}}\) for n going to infinity and i get \(\displaystyle \lim_{k\rightarrow +\infty}\frac{k+2}{k+1}=2\Rightarrow R=\frac{1}{2}\) but the real result should be 1. Am i doing something wrong? Thankyou

 

[Prove It]

Hello! I have a problem with the following exercise, in which i must calculate the ray of a power serie. This is the power serie: \(\displaystyle \sum_{K=0}^{+\infty}(k+1)z^{k+1}\). I decide to use the ratio test, and so i calculate \(\displaystyle \lim_{k\rightarrow +\infty}\frac{a_{n+1}}{a_{n}}\) for n going to infinity and i get \(\displaystyle \lim_{k\rightarrow +\infty}\frac{k+2}{k+1}=2\Rightarrow R=\frac{1}{2}\) but the real result should be 1. Am i doing something wrong? Thankyou

What is a "ray" of convergence? If you mean a "radius" or "circle" of convergence, then you are correct, you should use the ratio test...

$\displaystyle \begin{align*} \lim_{k \to \infty} \left| \frac{a_{k+1}}{a_k} \right| &< 1 \\ \lim_{k \to \infty} \left| \frac{ \left( k + 2 \right) \, z ^{k + 2} }{ \left( k + 1 \right) \, z^{k + 1} } \right| &< 1 \\ \lim_{k \to \infty} \left( \frac{ k + 2 }{ k + 1 } \right) \,\left| z \right| &< 1 \\ \lim_{ k \to \infty } \left( 1 + \frac{1}{k + 1} \right) \, \left| z \right| &< 1 \\ \left( 1 + 0 \right) \, \left| z \right| &< 1 \\ \left| z \right| &< 1 \end{align*}$

The radius of convergence is 1.

 

[math771]

Hi there!

It looks like you're on the right track with using the ratio test to calculate the radius of convergence for this power series. Your calculations seem correct, but there may be a small error in your final answer.

When taking the limit as k approaches infinity, you should be using n instead of k in your expression. So the correct calculation would be: \lim_{n\rightarrow +\infty}\frac{n+2}{n+1}=1.

This means that the radius of convergence is R=1, which matches the expected result. I hope this helps! Let me know if you have any other questions or if I can clarify anything for you. Good luck with your exercise!