quartz/content/Obsidian Vault/More tests for convergence.md

#math/calculus

Direct Comparison Test

\sum\limits \frac{1}{2^{n+1}} \textasciitilde \sum\limits \frac{1}{2^n}

1. if \sum b_{n} converges \&\ a_{n}<b_{n} \forall n => \sum a_{n} converges
2. if \sum b_{n} diverges \&\ a_{n}>b_{n} \forall n \Rightarrow \sum\limits a_{n} diverges

Limit Comparison Test

it's actually just an addition to the principle of an upper bound. It evaluates limits directly. for all \sum\limits a_{n}, \sum\limits b_{n} positive, \lim_{n\Rightarrow \infty} \frac{a_{n}}{b_{n}}=c exists if c>0 & is finite, then a follows b c=0 \Rightarrow both converge c=\infty \Rightarrow both diverge

Error for estimated

If the sum itself is hard to compute, bound it with a diverging series above and use that series' error function. R_{n}\leq T_{n} < \frac{1}{2n^{2}}