# How can we get Nth digit of any irrational number in base 2 using Taylor expansion?

For example:

$$\sqrt e = e^{0.5} = \sum^\infty_{n=0}\frac{(0.5^n)}{n!}$$

I know, we can easily get nth term of the series, but how can we get nth binary digit of that irrational number namely infinite sum of that infinite terms?

Or, are there more efficient approaches to get them?

For example, I want to get bits whose locations(orders) are in interval [100000-104096)

namely: 100000th bit, 104095th bit and every bits between them in order.

• The following might be useful. David H. Bailey and Richard E. Crandall. On the Random Character of Fundamental Constant Expansions. projecteuclid.org/euclid.em/999188630
– user63188
Feb 24, 2020 at 22:57

For example, with $$\pi=4\arctan 1$$, we can express it with the Taylor series, $$\arctan x=\sum\limits_{i=0}^\infty\frac{\left(-1\right)^{i}x^{2i+1}}{{2i+1}}$$, but so far, there is no known way to convert this to a "spigot algorithm" that returns a particular decimal of $$\pi$$. Such a result would be publishable but interestingly, the BBP formula is a spigot algorithm for base-16.