Xenaʻs Criterion
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Xenaʻs Criterion
This note is all about proving Xenaʻs Criterion as well as the troubleshooting method that comes with Xenaʻs Criterion. These methods show us when we have the right tools and cosmic experience to use Xenaʻs power. It also shows us what to do when even with the right tools and cosmic experience, the powers donʻt work. This is the troubleshooting part.
Taraʻs Criterion
Consider
\[X = \left\lbrace \frac{1}{2^n} \right\rbrace_{n = 0}^\infty = \left\lbrace1, \frac{1}{2}, \frac{1}{4},...\right\rbrace\]we know that $\inf\{X\} = 0$ and $\sup\{X\} = 1$. This is enough to guarantee convergence of the sequence. It is monotone decreasing since
\[X(n+1) < X(n) \iff \frac{1}{2^{n+1}} = \frac{1}{2}\left(\frac{1}{2^n}\right) < \frac{1}{2^n} \\ \frac{1}{2}\alpha < \alpha \ \ :\ \ \ [\alpha > 0], \ \alpha = 1/2\]From here, we can extrapolate since:
\[\frac{1}{2} > \frac{1}{\beta} \ \ \ \ \ (\beta \in \{3, 4, ...\})\]the family of sequences
\[X^{\beta} = \left\lbrace \frac{1}{\beta^n}\right\rbrace_{n=0}^\infty = \left\lbrace1, \frac{1}{\beta}, \frac{1}{\beta^2}, ...\right\rbrace\]is also monotone decreasing. Further, $\sup\left\{X^\beta\right\} = 1$ and $\inf\left\{X^\beta\right\} = 0$. From here, we know $X^\beta$ for $\beta \in \{2, 3, 4, …\}$ is convergent! This is as desired. $\blacksquare$
Taraʻs Troubleshooting
Suppose we have
\[\sum_{i = 0}^\infty \frac{1}{2^i} = 1 + \frac{1}{2} + \frac{1}{4} + \dots.\]We want to find out the exact convergence of this series. We will use the method of forms! So, we know
\[S_0 = 1 + \frac{1}{2} + \frac{1}{4} + \dots\]and if we take out a common factor, we get:
\[S_0 = 1 + \frac{1}{2}\left[1 + \frac{1}{2} + \frac{1}{4} + \dots\right] = 1 + \frac{1}{2}S_0\]Our goal is to find out what $S_0$ is. If we work with the algebra in the above equations we get…
\[S_0 = 1 + \frac{1}{2}S_0 \implies S_0 - \frac{1}{2}S_0 = 1\]This then turns out to be…
\[S_0\left[1 - \frac{1}{2}\right] = 1 \implies S_0 = 2\]So, then the solution by the method of forms is
\[\sum_{i = 0}^\infty \frac{1}{2^i} = 2\]as desired. $\blacksquare$
The pen is mightier than the sword!