Newton's evaluation of $1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$ How might have Newton evaluated the following series?
$$\sqrt{2} \, \frac{\pi}{4} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$$
The method of the this thread applies by setting $x=\pi/4$ in the Fourier series for $f(x) = \pi/2 - x/2$ and then subtracting the extraneous terms (which are a multiple of the Gregory-Leibniz series for $\pi/4$).
I read that this series appears in a letter from Newton to Leibniz.  However, I do not have access the letter which appears in this volume.

Although the question appears to be about how Newton historically did it, I'll convert a popular comment to an answer showing how techniques from his era, similar to those that handle the Gregory series, evaluate the series above:
$$\begin{align}\sum_{n\ge0}\left(\frac{1}{8n+1}+\frac{1}{8n+3}-\frac{1}{8n+5}-\frac{1}{8n+7}\right)&=\sum_{n\ge0}\int_{0}^{1}x^{8n}\left(1+x^{2}\right)\left(1-x^{4}\right)dx\\&=\int_{0}^{1}\frac{1+x^{2}}{1+x^{4}}dx\\&=\int_{0}^{1}\frac{1+x^{2}}{\left(1-x\sqrt{2}+x^{2}\right)\left(1+x\sqrt{2}+x^{2}\right)}dx\\&=\frac{1}{2}\sum_{\pm}\int_{0}^{1}\frac{dx}{1\pm x\sqrt{2}+x^{2}}\\&=\frac{1}{\sqrt{2}}\sum_{\pm}\left[\arctan\left(x\sqrt{2}\pm1\right)\right]_{0}^{1}\\&=\frac{\arctan\left(\sqrt{2}+1\right)+\arctan\left(\sqrt{2}-1\right)}{\sqrt{2}}\\&=\frac{\pi}{2\sqrt{2}}.\end{align}$$
