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| − | <span class="exam">For each the following series find the sum, if it converges. If | + | <span class="exam">For each the following series find the sum, if it converges. |
| − | you think it diverges, explain why.
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| − | ::<span class="exam">(a) (6 points) <math style="vertical-align: -50%">\frac{1}{2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3^{2}}-\frac{1}{2\cdot3^{3}}+\frac{1}{2\cdot3^{4}}-\frac{1}{2\cdot3^{5}}+\cdots .</math>
| + | <span class="exam">If you think it diverges, explain why. |
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| | + | <span class="exam">(a) <math style="vertical-align: -50%">\frac{1}{2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3^{2}}-\frac{1}{2\cdot3^{3}}+\frac{1}{2\cdot3^{4}}-\frac{1}{2\cdot3^{5}}+\cdots </math> |
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| − | ::<span class="exam">(b) (6 points) <math style="vertical-align: -75%"> \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}.</math>
| + | <span class="exam">(b) <math style="vertical-align: -75%"> \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}</math> |
| | + | <hr> |
| | + | [[009C Sample Midterm 3, Problem 2 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | ! Foundations:
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| − | |-
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| − | |One of the important series to know is the '''Geometric series.''' These are series with a common ratio <math style="vertical-align: 0%">r</math> between adjacent terms which are usually written
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| − | ::<math>\sum_{k=0}^{\infty}a_{0}r^{k}.</math>
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| − | |These are convergent if <math style="vertical-align: -22%">|r|<1</math>, and divergent if <math style="vertical-align: -22%">|r|\geq1</math>. If it is convergent, we can find the sum by the formula
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| − | ::<math>S=\frac{a_{0}}{1-r},</math><br>
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| − | |where <math style="vertical-align: -12%">a_{0}</math> is the first term in the series (if the index starts at <math style="vertical-align: 0%">k=2</math> or <math style="vertical-align: 0%">k=6</math>, then "<math style="vertical-align: -12%">a_{0}</math>" is actually the first term <math style="vertical-align: -12%">a_{2}</math> or <math style="vertical-align: -12%">a_{6}</math>, respectively).
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| − | |Another common type of series to evaluate is a '''telescoping series''', where the telescoping better describes the partial sums, denoted <math style="vertical-align: -15%">S_k</math>. Most of the time, they are presented as a fraction which requires '''partial fraction decomposition'''.
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| − | |This can be accomplished fairly quickly via a shortcut when the factors in the denominator are linear and share the same coefficient on <math style="vertical-align: 0%">n</math>.
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| − | |'''Example.''' Suppose we wish to decompose the fraction <math style="vertical-align: -22%">\frac{4}{(n-2)(n+1)}</math>. First, consider the difference
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| − | ::<math>\frac{1}{n-2}-\frac{1}{n-1}</math>
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| − | |If we combine this to a common denominator, we find
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| − | ::<math>\frac{1}{n-2}\cdot\frac{n+1}{n+1}-\frac{1}{n+1}\cdot\frac{n-2}{n-2}\,=\,\frac{n+1-(n-2)}{(n-2)(n+1)}\ =\ \frac{3}{(n-2)(n+1)}.</math>
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| − | |To have a 1 in the numerator, we would just multiply by <math style="vertical-align: -25%">1/3,</math> or the reciprocal of the difference between the two constants. Thus
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| − | ::<math>\begin{array}{rcl}
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| − | {\displaystyle \frac{4}{(n-2)(n+1)}} & = & {\displaystyle 4\cdot\frac{1}{(n-2)(n+1)}}\\
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| − | \\
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| − | & = & 4\cdot{\displaystyle \frac{1}{3}\left(\frac{1}{n-2}-\frac{1}{n+1}\right).}
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| − | \end{array}</math>
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| − | |Notice the pattern: for any fraction of the form <math style="vertical-align: -20%">\frac{1}{(x+a)(x+b)}</math> where <math style="vertical-align: -20%">a<b,</math> we have
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| − | ::<math>{\displaystyle \frac{1}{(x+a)(x+b)}\,=\,\frac{1}{b-a}\left(\frac{1}{x+a}-\frac{1}{x+b}\right).}</math>
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| − | |In this manner, we can quickly find that
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| − | ::<math>{\displaystyle \frac{1}{n^{2}-25}\,=\,\frac{1}{(n-5)(n+5)}\,=\,\frac{1}{5-(-5)}\left(\frac{1}{n-5}-\frac{1}{n+5}\right)\,=\,\frac{1}{10}\left(\frac{1}{n-5}-\frac{1}{n+5}\right)}</math>
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| − | |As per the so-called telescoping, consider the series defined by
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| − | ::<math>\sum_{n=2}^{\infty}\frac{1}{n^{2}-1}.</math>
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| − | |Using the technique above, we can rewrite the series as
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| − | ::<math>\sum_{n=2}^{\infty}\frac{1}{(n-1)(n+1)}\,=\,\sum_{n=2}^{\infty}\frac{1}{2}\left(\frac{1}{n-1}-\frac{1}{n+1}\right).</math>
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| − | |This means that
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| − | ::<math>\begin{array}{cclcl}
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| − | S_{2} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)\\
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| − | \\
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| − | S_{3} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}\right)\\
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| − | \\
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| − | S_{4} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{4}-\frac{1}{5}\right)\\
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| − | \\
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| − | S_{5} & = & \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{4}-\frac{1}{6}\right) & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{5}-\frac{1}{6}\right)\\
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| − | \\
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| − | \vdots & \vdots & \vdots\\
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| − | S_{k} & = & \frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{k}-\frac{1}{k+1}\right).
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| − | \end{array}</math>
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| − | |Again, notice the pattern: each time there are ''<u>exactly</u>'' two surviving positive terms, and two surviving negative terms in each partial sum. If we then take the limit, we find
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| − | ::<math>\sum_{n=2}^{\infty}\frac{1}{n^{2}-1}\,=\,\lim_{n\rightarrow\infty}S_{n}\,=\,\lim_{n\rightarrow\infty}\frac{1}{2}\left(\frac{1}{1}+\frac{1}{2}-\frac{1}{n}-\frac{1}{n+1}\right)\,=\,\frac{3}{4}.</math>
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| − | |}
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| − | '''Solution:'''
| + | [[009C Sample Midterm 3, Problem 2 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(a):
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| − | |This is the easier portion of the problem. Each term grows by a ratio of <math style="vertical-align: -22%">1/3</math>, and it reverses sign. Thus, there is a common ratio <math style="vertical-align: -22%">r=-1/3</math>. Also, the first term is <math style="vertical-align: -22%">1/2</math>, so we can write the series as a geometric series:
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| − | ::<math style="vertical-align: -75%"> \sum_{n=0}^{\infty}\,\frac{1}{2}\left(-\frac{1}{3}\right)^n.</math>
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| − | |Then, the series converges to the sum
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| − | ::<math>S\,=\,\frac{a_{0}}{1-r}\,=\,\frac {1/2}{1-(-1/3)}\,=\,\frac{1/2}{4/3}\,=\,\frac{3}{8}.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !(b):
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| − | |Using the technique in Foundations, we can rewrite the sequence as
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| − | ::<math> \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}\,=\,3 \sum_{n=1}^{\infty}\,\frac{1}{2} \left(\frac{1}{2n-1}-\frac{1}{2n+1}\right).</math>
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| − | Writing a few terms out, we find
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| − | ::<math>S\,=\,\frac{3}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\cdots \right).</math>
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| − | |Since only one positive term and one negative term survive, we have
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| − | |<math> S_k \,=\,\frac{3}{2}\left( 1-\frac{1}{2k+1} \right),</math>
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| − | |so the series converges to the sum ''S'', where
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| − | ::<math>S\,=\,\lim_{k\rightarrow\infty} S_k\,=\,\frac{3}{2}.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | |The series in (a) converges to <math style="vertical-align: -21%">3/8</math>, while the series in (b) converges to <math style="vertical-align: -21%">3/2</math>.
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| − | |}
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| | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] | | [[009C_Sample_Midterm_3|'''<u>Return to Sample Exam</u>''']] |
