sin[(2n-1)pi]/2

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- Oct 23rd 2012, 06:42 PMhiggins19Alternating Series test
sin[(2n-1)pi]/2

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