Series Expansion for Pi over Root 2/Mistake

Source Work

• 1961: I.N. Sneddon: Fourier Series: Exercises on Chapter $\text I$: $2$.

Mistake

Deduce that

$\ds \sum_{n \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} } = \frac \pi {\sqrt 2}$

Correction

That lower index should of course be $r$:

$\ds \sum_{r \mathop = 1}^\infty \paren {-1}^{r - 1} \frac {r - \frac 1 2} {r^2 - r + \frac 3 {16} } = \frac \pi {\sqrt 2}$

otherwise it makes no sense.

Sources

• 1961: I.N. Sneddon: Fourier Series ... (previous) ... (next): Exercises on Chapter $\text I$: $2$.