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.

Looking at the text mentioned, we see something interesting on page $156$, in note $(48)$. Transcribing it (you can find a screencap of the text here):

On observing that $$1 + \frac 1 3 - \frac 1 5 - \frac 1 7 + \frac 1 9 + \text{etc.} = \int_0^1 \frac{1+x^2}{1+x^4}dx$$ by expanding the integrand as an ascending series in $x$, we may suppose that Newton obtained his result by integrating the identity $$\frac{1}{e+fz+gz^2} + \frac{1}{e-fz+gz^2} = \frac{2e + 2gz^2}{e^2 + g^2 z^4} \text{ (on putting } 2eg = f^2 \text{)}$$ in two ways. Put $z \sqrt g = x \sqrt e$ and $-x\sqrt e$, respectively, in the two fractions on the left, and integrate from $0$ to $1$ with regard to $x$. Then, on combining the terms on the left, the identity gives $$\frac 1 2 \int_{-1}^1 \frac{dx}{1 + \sqrt 2 x + x^2} = \int_0^1 \frac{1+x^2}{1+x^4}dx$$ By putting $1 + x \sqrt 2 = \tan \theta$, and integrating $\theta$ from $- \frac 1 8 \pi$ to $\frac 3 8 \pi$, the left side gives $\pi/2 \sqrt 2$, and the right is equal to the series. (Cf. Hofmann, p. $175$.) The series, when written $1 + (\frac 1 3 - \frac 1 5) - \frac 1 7 - \frac 1 9) + \text{etc.}$, leads to the next result in the letter.

Granted, I'm not sure if this is how Newton actually calculated it, and I haven't actually read this book thoroughly enough to say if it's what you're looking for (I just skimmed until I found what seemed relevant). Hopefully it's enlightening though.
