You signed in with another tab or window. Reload to refresh your session.You signed out in another tab or window. Reload to refresh your session.You switched accounts on another tab or window. Reload to refresh your session.Dismiss alert
### nth Term Test for Divergence nth-Term Test for Divergence
@@ -118,6 +129,57 @@ $$ \text {2. } a_{n-1} \le a_n \text{ for all } n $$
118
129
119
130
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.
120
131
132
+
### Alternating Series Remainder
133
+
134
+
Todo: review as needed c.f. L&E page 635.
135
+
121
136
### Absolute Convergence
122
137
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$,
<h2>Tests for Convergence and Divergence<aclass="headerlink" href="#tests-for-convergence-and-divergence" title="Link to this heading"></a></h2>
@@ -203,8 +211,48 @@ <h3>Alternating Series Test<a class="headerlink" href="#alternating-series-test"
203
211
\[ \text {2. } a_{n-1} \le a_n \text{ for all } n \]</div>
204
212
<p>To prove the second point, that the function is descending everywhere, we take this approach: let <spanclass="math notranslate nohighlight">\(a_n = f(x)\)</span>, then we can show that <spanclass="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>
205
213
</section>
214
+
<sectionid="alternating-series-remainder">
215
+
<h3>Alternating Series Remainder<aclass="headerlink" href="#alternating-series-remainder" title="Link to this heading"></a></h3>
216
+
<p>Todo: review as needed c.f. L&E page 635.</p>
217
+
</section>
206
218
<sectionid="absolute-convergence">
207
219
<h3>Absolute Convergence<aclass="headerlink" href="#absolute-convergence" title="Link to this heading"></a></h3>
220
+
<p>Some series with different signs do not strictly alternate, such as:</p>
221
+
<divclass="math notranslate nohighlight">
222
+
\[ \sum_{n=1}^\infty \frac{\sin n}{n^2} \]</div>
223
+
<p>We can see if the series of the terms absolute values converge – if so, it is “absolutely convergent.” In this case, since <spanclass="math notranslate nohighlight">\( | \sin n| \le 1 \)</span> for all n, for all <spanclass="math notranslate nohighlight">\( n > 1\)</span>,</p>
<p>Therefore the series converges. This means that the series without the absolute value converges.</p>
227
+
<p>Absolute convergence is the stronger claim.</p>
228
+
<p>Theorems and definitions:</p>
229
+
<p>If the series <spanclass="math notranslate nohighlight">\(\sum \space |a_n|\)</span> converges, then the series <spanclass="math notranslate nohighlight">\(\sum a_n\)</span> converges. This is absolute convergence.</p>
230
+
<p>If the series <spanclass="math notranslate nohighlight">\(\sum \space |a_n|\)</span> does not converge but the series <spanclass="math notranslate nohighlight">\(\sum a_n\)</span> does, the series is conditionally convergent.</p>
231
+
</section>
232
+
<sectionid="the-ratio-test-and-root-test">
233
+
<h3>The Ratio Test and Root Test<aclass="headerlink" href="#the-ratio-test-and-root-test" title="Link to this heading"></a></h3>
234
+
<sectionid="the-ratio-test">
235
+
<h4>The Ratio Test<aclass="headerlink" href="#the-ratio-test" title="Link to this heading"></a></h4>
236
+
<p>Let <spanclass="math notranslate nohighlight">\(\sum a_n\)</span> be a series with non-zero terms.</p>
### nth Term Test for Divergence nth-Term Test for Divergence
@@ -118,6 +129,57 @@ $$ \text {2. } a_{n-1} \le a_n \text{ for all } n $$
118
129
119
130
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.
120
131
132
+
### Alternating Series Remainder
133
+
134
+
Todo: review as needed c.f. L&E page 635.
135
+
121
136
### Absolute Convergence
122
137
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$,
0 commit comments