Can a series $\sum_{n=1}^{\infty} (\frac{1}2)^n=1$ be infinitely substituted into itself

algebra-precalculussequences-and-series

Example

$$\Large\sum_{n=1}^{\infty} \left(\frac{\sum_{n=1}^{\infty} \left(\frac{\sum_{n=1}^{\infty} \left(\frac{\sum_{n=1}^{\infty} \left(\frac{\sum_{n=1}^{\infty} \left(\frac{…}2\right)^n}2\right)^n}2\right)^n}2\right)^n}2\right)^n$$

Motivation

Recently did a screen share on gather.town with itself and was greeted with this image below. The idea popped into my head, "What if this infinite mirror concept was applied to an infinite series, would it be possible to create a series that has a known result but is not calculable?"

infinity_mirror

Best Answer

What you end up with is the relation

$$x = \sum_{n=1}^\infty \left(\frac{x}{2}\right)^n.$$

If we assume $x$ to be complex we must have $|x| < 2$ or the RHS diverges. Then, summing the geometric series

$$x = -1 + \sum_{n=0}^\infty \left(\frac{x}{2}\right)^n$$ $$x = -1 + \frac{1}{1-\frac{x}{2}}$$ $$-\frac{1}{2}x^2 +\frac{1}{2}x = 0$$

gives $x = 0 \vee x = 1$ as the only solutions.

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