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20 lines
811 B
Markdown
20 lines
811 B
Markdown
#math/calculus
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# Direct Comparison Test
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$\sum\limits \frac{1}{2^{n+1}} \textasciitilde \sum\limits \frac{1}{2^n}$
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1. if $\sum b_{n}$ converges $\&\ a_{n}<b_{n} \forall n => \sum a_{n}$ converges
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2. if $\sum b_{n}$ diverges $\&\ a_{n}>b_{n} \forall n \Rightarrow \sum\limits a_{n}$ diverges
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# Limit Comparison Test
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it's actually just an addition to the principle of an upper bound. It evaluates limits directly.
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for all $\sum\limits a_{n}, \sum\limits b_{n}$ positive,
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$\lim_{n\Rightarrow \infty} \frac{a_{n}}{b_{n}}=c$ exists
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if c>0 & is finite, then a follows b
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$c=0 \Rightarrow$ both converge
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$c=\infty \Rightarrow$ both diverge
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# Error for estimated
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If the sum itself is hard to compute, bound it with a diverging series above and use that series' error function.
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$R_{n}\leq T_{n} < \frac{1}{2n^{2}}$ |