# Alternating Series test

• Oct 23rd 2012, 06:42 PM
higgins19
Alternating Series test
sin[(2n-1)pi]/2
• Oct 23rd 2012, 07:01 PM
Prove It
Re: Alternating Series test
Do you really need to use the alternating series test here? You should know that \displaystyle \begin{align*} \sin{\left[ \frac{\left( 2n - 1 \right) \pi}{2} \right]} = (-1)^{n + 1} \end{align*}, and so adding them, the series can't possibly converge because it can only ever take the values of 1 and 0.