QUESTION: Interval of Convergens for a series

[Hummingbird25]

Hi

I have this series here

\sum_{n=1} ^{\infty} \frac{1}{x^2+n^2}

I need to show that the Radius of convergens R = \infty and the interval of convergens therefore is (- \infty, \infty)

My question is to do this don't I use the ratio-test?

Sincerely Yours

Hummingbird25

 

[eok20]

a_n is the nth term, so:

\left| \frac{a_{n+1}}{a_n} \right | = \left| \frac{1}{x^2+(n+1)^2} \cdot \frac{x^2+n^2}{1} \right |

However, this goes to 1 as n goes to inifinity so this really doesn't help you. What you probably want to do is use comparison test with 1/n^2 to show that it converges for any x.

 

[Hummingbird25]

Hi and thanks for Your answer,

\sum_{n=1} ^{\infty} \frac{1}{x^2+n^2}

Then by the comparison test:

\frac{1}{x^2 + n^2} < \frac{1}{n^2} ??

Sincerely

Hummingbird25

a_n is the nth term, so:

\left| \frac{a_{n+1}}{a_n} \right | = \left| \frac{1}{x^2+(n+1)^2} \cdot \frac{x^2+n^2}{1} \right |

However, this goes to 1 as n goes to inifinity so this really doesn't help you. What you probably want to do is use comparison test with 1/n^2 to show that it converges for any x.

 

[eok20]

Then by the comparison test:

\frac{1}{x^2 + n^2} < \frac{1}{n^2} ??

Thats right, and since \sum_{n=1} ^{\infty} \frac{1}{n^2} converges (by p-series), \sum_{n=1} ^{\infty} \frac{1}{x^2+n^2} converges since every term is smaller.