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Update sequences and series notes
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docs/.doctrees/environment.pickle

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docs/_sources/sequences-and-series-notes.md.txt

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@@ -28,6 +28,17 @@ The term $n=1$ at the bottom is our starting point, the 3 on the top is our endp
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## Properties of Infinite Series (p. 612)
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$
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\begin{aligned} \sum_{n=1}^\infty c \cdot a_n = c \cdot \sum_{n=1}^\infty a \end{aligned} \\
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$
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$
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\begin {aligned}
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\sum_{n=1}^\infty (a_n \pm b_n) = \sum_{n=1}^\infty a_n \pm \sum_{n=1}^\infty b_n
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\end{aligned}
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$
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## Tests for Convergence and Divergence
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### nth Term Test for Divergence nth-Term Test for Divergence
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To prove the second point, that the function is descending everywhere, we take this approach: let $a_n = f(x)$, then we can show that $\frac{d}{dx} f(x)$ is negative for the domain we're considering. Remember ascending and descending functions from Calc 1.
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### Alternating Series Remainder
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Todo: review as needed c.f. L&E page 635.
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### Absolute Convergence
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Some series with different signs do not strictly alternate, such as:
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$$ \sum_{n=1}^\infty \frac{\sin n}{n^2} $$
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We can see if the series of the terms absolute values converge -- if so, it is "absolutely convergent." In this case, since $ | \sin n| \le 1 $ for all n, for all $ n > 1$,
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$$ \left| \frac{\sin n}{n^2} \right| <= \frac{1}{n^2} $$
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Therefore the series converges. This means that the series without the absolute value converges.
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Absolute convergence is the stronger claim.
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Theorems and definitions:
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If the series $\sum \space |a_n|$ converges, then the series $\sum a_n$ converges. This is absolute convergence.
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If the series $\sum \space |a_n|$ does not converge but the series $\sum a_n$ does, the series is conditionally convergent.
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### The Ratio Test and Root Test
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#### The Ratio Test
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Let $\sum a_n$ be a series with non-zero terms.
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Then:
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$
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| < 1 \text{ converges absolutely.}\\
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\end{aligned}
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$
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$
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| > 1 \space (\text{or } \infty ) \text{ diverges.}\\
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\end{aligned}
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$
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$
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| = 1 \text{ is inconclusive.}\\
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\end{aligned}$
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#### The Root Test
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Not as useful as the ratio test, TODO.
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docs/objects.inv

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docs/searchindex.js

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docs/sequences-and-series-notes/index.html

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</ul>
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<p class="caption" role="heading"><span class="caption-text">Added Recently:</span></p>
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<ul>
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<li class="toctree-l1"><a class="reference internal" href="../complex-inner-products/">The Inner Product of a Complex Vectors</a></li>
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<li class="toctree-l1"><a class="reference internal" href="../complex-inner-products/">The Inner Product of Complex Vectors</a></li>
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<li class="toctree-l1"><a class="reference internal" href="../getting-started-with-ollama/">Getting Started With Ollama</a></li>
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<li class="toctree-l1"><a class="reference internal" href="../math/yet-another-jdesmos-demo/">Yet Another JDesmos Demo</a></li>
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<li class="toctree-l1"><a class="reference internal" href="../math/desmos-jupyter/">Using Desmos Graphs in Jupyter Notebooks</a></li>
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</section>
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<section id="properties-of-infinite-series-p-612">
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<h2>Properties of Infinite Series (p. 612)<a class="headerlink" href="#properties-of-infinite-series-p-612" title="Link to this heading"></a></h2>
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<p><span class="math notranslate nohighlight">\(
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\begin{aligned} \sum_{n=1}^\infty c \cdot a_n = c \cdot \sum_{n=1}^\infty a \end{aligned} \\
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\)</span></p>
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<p><span class="math notranslate nohighlight">\(
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\begin {aligned}
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\sum_{n=1}^\infty (a_n \pm b_n) = \sum_{n=1}^\infty a_n \pm \sum_{n=1}^\infty b_n
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\end{aligned}
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\)</span></p>
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</section>
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<section id="tests-for-convergence-and-divergence">
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<h2>Tests for Convergence and Divergence<a class="headerlink" href="#tests-for-convergence-and-divergence" title="Link to this heading"></a></h2>
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\[ \text {2. } a_{n-1} \le a_n \text{ for all } n \]</div>
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<p>To prove the second point, that the function is descending everywhere, we take this approach: let <span class="math notranslate nohighlight">\(a_n = f(x)\)</span>, then we can show that <span class="math notranslate nohighlight">\(\frac{d}{dx} f(x)\)</span> is negative for the domain we’re considering. Remember ascending and descending functions from Calc 1.</p>
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</section>
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<section id="alternating-series-remainder">
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<h3>Alternating Series Remainder<a class="headerlink" href="#alternating-series-remainder" title="Link to this heading"></a></h3>
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<p>Todo: review as needed c.f. L&amp;E page 635.</p>
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</section>
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<section id="absolute-convergence">
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<h3>Absolute Convergence<a class="headerlink" href="#absolute-convergence" title="Link to this heading"></a></h3>
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<p>Some series with different signs do not strictly alternate, such as:</p>
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<div class="math notranslate nohighlight">
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\[ \sum_{n=1}^\infty \frac{\sin n}{n^2} \]</div>
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<p>We can see if the series of the terms absolute values converge – if so, it is “absolutely convergent.” In this case, since <span class="math notranslate nohighlight">\( | \sin n| \le 1 \)</span> for all n, for all <span class="math notranslate nohighlight">\( n &gt; 1\)</span>,</p>
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<div class="math notranslate nohighlight">
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\[ \left| \frac{\sin n}{n^2} \right| &lt;= \frac{1}{n^2} \]</div>
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<p>Therefore the series converges. This means that the series without the absolute value converges.</p>
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<p>Absolute convergence is the stronger claim.</p>
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<p>Theorems and definitions:</p>
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<p>If the series <span class="math notranslate nohighlight">\(\sum \space |a_n|\)</span> converges, then the series <span class="math notranslate nohighlight">\(\sum a_n\)</span> converges. This is absolute convergence.</p>
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<p>If the series <span class="math notranslate nohighlight">\(\sum \space |a_n|\)</span> does not converge but the series <span class="math notranslate nohighlight">\(\sum a_n\)</span> does, the series is conditionally convergent.</p>
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</section>
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<section id="the-ratio-test-and-root-test">
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<h3>The Ratio Test and Root Test<a class="headerlink" href="#the-ratio-test-and-root-test" title="Link to this heading"></a></h3>
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<section id="the-ratio-test">
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<h4>The Ratio Test<a class="headerlink" href="#the-ratio-test" title="Link to this heading"></a></h4>
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<p>Let <span class="math notranslate nohighlight">\(\sum a_n\)</span> be a series with non-zero terms.</p>
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<p>Then:</p>
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<p><span class="math notranslate nohighlight">\(
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| &lt; 1 \text{ converges absolutely.}\\
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\end{aligned}
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\)</span></p>
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<p><span class="math notranslate nohighlight">\(
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| &gt; 1 \space (\text{or } \infty ) \text{ diverges.}\\
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\end{aligned}
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\)</span></p>
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<p><span class="math notranslate nohighlight">\(
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| = 1 \text{ is inconclusive.}\\
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\end{aligned}\)</span></p>
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</section>
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<section id="the-root-test">
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<h4>The Root Test<a class="headerlink" href="#the-root-test" title="Link to this heading"></a></h4>
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<p>Not as useful as the ratio test, TODO.</p>
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<hr class="footnotes docutils" />
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<aside class="footnote-list brackets">
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<aside class="footnote brackets" id="id2" role="note">
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</aside>
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</section>
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</section>
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</section>
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</section>
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src/sequences-and-series-notes.md

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@@ -28,6 +28,17 @@ The term $n=1$ at the bottom is our starting point, the 3 on the top is our endp
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## Properties of Infinite Series (p. 612)
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$
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\begin{aligned} \sum_{n=1}^\infty c \cdot a_n = c \cdot \sum_{n=1}^\infty a \end{aligned} \\
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$
34+
35+
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$
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\begin {aligned}
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\sum_{n=1}^\infty (a_n \pm b_n) = \sum_{n=1}^\infty a_n \pm \sum_{n=1}^\infty b_n
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\end{aligned}
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$
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## Tests for Convergence and Divergence
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### nth Term Test for Divergence nth-Term Test for Divergence
@@ -118,6 +129,57 @@ $$ \text {2. } a_{n-1} \le a_n \text{ for all } n $$
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To prove the second point, that the function is descending everywhere, we take this approach: let $a_n = f(x)$, then we can show that $\frac{d}{dx} f(x)$ is negative for the domain we're considering. Remember ascending and descending functions from Calc 1.
120131

132+
### Alternating Series Remainder
133+
134+
Todo: review as needed c.f. L&E page 635.
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121136
### Absolute Convergence
122137

138+
Some series with different signs do not strictly alternate, such as:
139+
140+
$$ \sum_{n=1}^\infty \frac{\sin n}{n^2} $$
141+
142+
We can see if the series of the terms absolute values converge -- if so, it is "absolutely convergent." In this case, since $ | \sin n| \le 1 $ for all n, for all $ n > 1$,
143+
144+
$$ \left| \frac{\sin n}{n^2} \right| <= \frac{1}{n^2} $$
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Therefore the series converges. This means that the series without the absolute value converges.
147+
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Absolute convergence is the stronger claim.
149+
150+
Theorems and definitions:
151+
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If the series $\sum \space |a_n|$ converges, then the series $\sum a_n$ converges. This is absolute convergence.
153+
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If the series $\sum \space |a_n|$ does not converge but the series $\sum a_n$ does, the series is conditionally convergent.
155+
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### The Ratio Test and Root Test
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#### The Ratio Test
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Let $\sum a_n$ be a series with non-zero terms.
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Then:
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$
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| < 1 \text{ converges absolutely.}\\
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\end{aligned}
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$
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$
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| > 1 \space (\text{or } \infty ) \text{ diverges.}\\
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\end{aligned}
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$
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$
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\begin{aligned}
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\lim \limits_{n \to \infty} \left| \frac{a_n + 1}{a_n} \right| = 1 \text{ is inconclusive.}\\
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\end{aligned}$
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#### The Root Test
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Not as useful as the ratio test, TODO.
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123185

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