|
|
| (One intermediate revision by the same user not shown) |
| Line 1: |
Line 1: |
| − | <span class="exam">Test if each the following series converges or diverges. Give reasons | + | <span class="exam">Test if each the following series converges or diverges. |
| − | and clearly state if you are using any standard test.
| |
| | | | |
| − | ::<span class="exam">(a) (6 points) <math>{\displaystyle \sum_{n=1}^{\infty}}\,\frac{n!}{(3n+1)!}.</math>
| + | <span class="exam">Give reasons and clearly state if you are using any standard test. |
| | + | |
| | + | <span class="exam">(a) <math>{\displaystyle \sum_{n=1}^{\infty}}\,\frac{n!}{(3n+1)!}</math> |
| | <br> | | <br> |
| | | | |
| − | ::<span class="exam">(b) (6 points) <math>{\displaystyle \sum_{n=2}^{\infty}}\,\frac{\sqrt{n}}{n^{2}-3}.</math>
| + | <span class="exam">(b) <math>{\displaystyle \sum_{n=2}^{\infty}}\,\frac{\sqrt{n}}{n^{2}-3}</math> |
| − | | + | <hr> |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Midterm 3, Problem 3 Solution|'''<u>Solution</u>''']] |
| − | ! Foundations:
| |
| − | |-
| |
| − | |Most of the time, if there are factorials in the terms of a series, you would use the
| |
| − | |-
| |
| − | |'''Ratio Test.''' Let <math style="vertical-align: -98%">\sum_{k=1}^{\infty} a_{k}</math> be a series. Then:
| |
| − | |-
| |
| − | |
| |
| − | :*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L<1</math>, the series is absolutely convergent (and therefore convergent).
| |
| − | |-
| |
| − | |
| |
| − | :*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L>1</math> or <math style="vertical-align: -83%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=\infty</math>, the series is divergent.
| |
| − | |-
| |
| − | |
| |
| − | :*If <math style="vertical-align: -84%">\lim_{k\rightarrow\infty}\left|\frac{a_{k+1}}{a_{k}}\right|=L=1</math>, the Ratio Test is inconclusive.
| |
| − | |-
| |
| − | |This works well, as factorials cancel out many terms. For example,
| |
| − | |-
| |
| − | |
| |
| − | ::<math>\frac{(n+1)!}{n!}\,=\,n+1.</math>
| |
| − | |-
| |
| − | |On the other hand, something built mainly out of powers of <math style="vertical-align: 0%">n</math> may work well with the | |
| − | |-
| |
| − | |'''Limit Comparison Test.''' Suppose <math style="vertical-align: -100%">\sum_{k=1}^{\infty} a_{k}</math> and <math style="vertical-align: -100%">\sum_{k=1}^{\infty} b_{k}</math> are series with positive terms. If <math style="vertical-align: -75%">\lim_{k\rightarrow\infty}\frac{a_{k}}{b_{k}}=c</math> where <math style="vertical-align: -5%">0<c<\infty</math>, then either both series converge, or both series diverge.
| |
| − | |-
| |
| − | |In the case of a series built mainly out of powers, you would choose to compare it to a ''p''-series.
| |
| − | |}
| |
| − | | |
| − | '''Solution:'''
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !(a):
| |
| − | |-
| |
| − | |As mentioned in Foundations, we should use the ratio test. Note that
| |
| − | |-
| |
| − | |
| |
| − | ::<math style="vertical-align: 0%">\begin{array}{rcl}
| |
| − | \displaystyle {\left|\frac{a_{n+1}}{a_{n}}\right|} & = & \left|\frac{\frac{(n+1)!}{(3(n+1)+1)!}}{\frac{n!}{(3n+1)!}}\right|\\
| |
| − | \\
| |
| − | & = & \displaystyle {\frac{(n+1)!}{n!}\cdot\frac{(3n+1)!}{(3n+4)!}}\\
| |
| − | \\
| |
| − | & = & \displaystyle {\frac{n+1}{(3n+2)(3n+3)(3n+4)}}.
| |
| − | \end{array}</math>
| |
| − | |-
| |
| − | |Thus,
| |
| − | |-
| |
| − | |
| |
| − | ::<math style="vertical-align: 0%">\lim_{n\rightarrow\infty}\left|\frac{a_{n+1}}{a_{n}}\right|\ =\ 0\ <\ 1,</math>
| |
| − | |-
| |
| − | |so by the ratio test the series converges.
| |
| − | |}
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | [[009C Sample Midterm 3, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| − | !(b):
| |
| − | |-
| |
| − | |Here, we can use the limit comparison test. Let <math style="vertical-align: -62%">a_{n}=\frac{\sqrt{n}}{n^{2}-3}</math>, and let <math style="vertical-align: -62%">b_{n}=\frac{1}{n^{3/2}}.</math> Notice that the terms of <math style="vertical-align: -10%">a_{n}</math> are all positive, and
| |
| − | |-
| |
| − | | | |
| − | ::<math style="vertical-align: 0%">\begin{array}{rcl}
| |
| − | \displaystyle {\lim_{n\rightarrow\infty}\frac{a_{n}}{b_{n}}} & = & \displaystyle {\lim_{n\rightarrow\infty}\frac{\frac{\sqrt{n}}{n^{2}-3}}{\frac{1}{n^{3/2}}}}\\
| |
| − | \\
| |
| − | & = & \displaystyle {\lim_{n\rightarrow\infty}\frac{n^{2}}{n^{2}-3}}\\
| |
| − | \\
| |
| − | & = & 1\,\,<\,\, \infty.
| |
| − | \end{array}</math>
| |
| | | | |
| − | Since <math style="vertical-align: -90%">\sum_{n=2}^{\infty}\frac{1}{n^{3/2}}</math>  is a p-series with <math style="vertical-align: -20%">p>1,</math>
| |
| − | it is convergent. By the limit comparison test, <math style="vertical-align: -90%">\sum_{n=2}^{\infty}\frac{\sqrt{n}}{n^{2}-3}</math>
| |
| − |  is convergent.
| |
| | | | |
| − | |}
| |
| | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
