Series Expansion for Pi over Root 2/Mistake

Source Work

 * : Exercises on Chapter $\text I$: $2$.

Mistake

 * Deduce that
 * $\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} } = \frac \pi {\sqrt 2}$

That lower index should of course be $r$:


 * $\displaystyle \sum_{r \mathop = 1}^\infty \left({-1}\right)^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} } = \frac \pi {\sqrt 2}$

otherwise it makes no sense.