Series with Hyperbolic and Trigonometric functions

azatkgz
Messages
182
Reaction score
0

Homework Statement


Determine whether the series converges and diverges.

\sum_{n=3}^{\infty}\ln \left(\frac{\cosh \frac{\pi}{n}}{\cos \frac{\pi}{n}}\right)






The Attempt at a Solution



\sum_{n=3}^{\infty}\ln \left(\frac{1+\frac{\pi^2}{2n^2}+O(\frac{1}{n^4})}{1-\frac{\pi^2}{2n^2}+O(\frac{1}{n^4})}\right)

=\sum_{n=3}^{\infty}\ln \left(\left(1+\frac{\pi^2}{2n^2}+O(\frac{1}{n^4})\right)\left(1+\frac{\pi^2}{2n^2}+O(\frac{1}{n^4})\right)\right)=\sum_{n=3}^{\infty}\ln \left(1+\frac{\pi^2}{n^2}+O(\frac{1}{n^4})\right)

=\sum_{n=3}^{\infty}\left(\frac{\pi^2}{n^2}+O(\frac{1}{n^4})\right)

series converges
 
Physics news on Phys.org
Sorry its not immediately obvious to me how you got the your first line of working to your second.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
Back
Top