009C Sample Midterm 3, Problem 2 Detailed Solution
For each the following series find the sum, if it converges.
If you think it diverges, explain why.
(a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \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 }
(b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=1}^{\infty}\,\frac{3}{(2n-1)(2n+1)}}
| Background Information: |
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| 1. For a geometric series Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty} ar^n} with Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle |r|<1,} |
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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sum_{n=0}^{\infty} ar^n=\frac{a}{1-r}.} |
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2. For a telescoping series, we find the sum by first looking at the partial sum Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_k} |
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and then calculate Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \lim _{k\rightarrow \infty }s_{k}.} |
Solution:
(a)
| Step 1: |
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| Each term grows by a ratio of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{3}} and it reverses sign. |
| Thus, there is a common ratio Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle r=-\frac{1}{3}.} |
| Also, the first term is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}.} So, we can write the series as a geometric series given by |
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| Step 2: |
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| Then, the series converges to the sum |
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Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{S} & = & \displaystyle{\frac{a}{1-r}}\\ &&\\ & = & \displaystyle{\frac {\frac{1}{2}}{1-(-\frac{1}{3})}}\\ &&\\ & = & \displaystyle{\frac{\frac{1}{2}}{\frac{4}{3}}}\\ &&\\ & = & \displaystyle{\frac{3}{8}.} \end{array}} |
(b)
| Step 1: |
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| We begin by using partial fraction decomposition. Let |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3}{(2x-1)(2x+1)}=\frac{A}{2x-1}+\frac{B}{2x+1}.} |
| If we multiply this equation by Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (2x-1)(2x+1),} we get |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3=A(2x+1)+B(2x-1).} |
| If we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=\frac{1}{2},} we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A=\frac{3}{2}.} |
| If we let Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x=-\frac{1}{2},} we get Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle B=-\frac{3}{2}.} |
| So, we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\sum_{n=1}^\infty \frac{3}{(2n-1)(2n+1)}} & = & \displaystyle{\sum_{n=1}^\infty \frac{\frac{3}{2}}{2n-1}+\frac{-\frac{3}{2}}{2n+1}}\\ &&\\ & = & \displaystyle{\frac{3}{2} \sum_{n=1}^\infty \frac{1}{2n-1}-\frac{1}{2n+1}.} \end{array}} |
| Step 2: |
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| Now, we look at the partial sums, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_n} of this series. |
| First, we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_1=\frac{3}{2}\bigg(1-\frac{1}{3}\bigg).} |
| Also, we have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{s_2} & = & \displaystyle{\frac{3}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}\bigg)}\\ &&\\ & = & \displaystyle{\frac{3}{2}\bigg(1-\frac{1}{5}\bigg)} \end{array}} |
| and |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{s_3} & = & \displaystyle{\frac{3}{2}\bigg(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}\bigg)}\\ &&\\ & = & \displaystyle{\frac{3}{2}\bigg(1-\frac{1}{7}\bigg).} \end{array}} |
| If we compare Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_1,s_2,s_3,} we notice a pattern. |
| We have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle s_n=\frac{3}{2}\bigg(1-\frac{1}{2n+1}\bigg).} |
| Step 3: |
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| Now, to calculate the sum of this series we need to calculate |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \lim_{n\rightarrow \infty} s_n.} |
| We have |
| Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\lim_{n\rightarrow \infty} s_n} & = & \displaystyle{\lim_{n\rightarrow \infty} \frac{3}{2}\bigg(1-\frac{1}{2n+1}\bigg)}\\ &&\\ & = & \displaystyle{\frac{3}{2}.} \end{array}} |
| Since the partial sums converge, the series converges and the sum of the series is Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3}{2}.} |
| Final Answer: |
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| (a) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3}{8}} |
| (b) Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{3}{2}} |