Difference between revisions of "009C Sample Final 1, Problem 2"
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| − | '''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math> | + | '''1.''' For a geometric series <math>\sum_{n=0}^{\infty} ar^n</math> with <math>|r|<1,</math> |
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| − | '''2.''' For a telescoping series, we find the sum by first looking at the partial sum <math style="vertical-align: -3px">s_k</math> | + | '''2.''' For a telescoping series, we find the sum by first looking at the partial sum <math style="vertical-align: -3px">s_k</math> |
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| − | |Since <math style="vertical-align: -16px">2<e,~\bigg|-\frac{2}{e}\bigg|<1.</math> So, | + | |Since <math style="vertical-align: -16px">2<e,~\bigg|-\frac{2}{e}\bigg|<1.</math> |
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| + | |So, | ||
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|This is a telescoping series. First, we find the partial sum of this series. | |This is a telescoping series. First, we find the partial sum of this series. | ||
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| − | |Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math> | + | |Let <math style="vertical-align: -20px">s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).</math> |
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|Then, | |Then, | ||
Revision as of 14:57, 26 February 2017
Find the sum of the following series:
(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 \sum_{n=0}^{\infty} (-2)^ne^{-n}}
(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} \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)}
| Foundations: |
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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} |
|
and then 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_{k\rightarrow\infty} s_k.} |
Solution:
(a)
| Step 1: |
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| First, we write |
|
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=0}^{\infty} (-2)^n e^{-n}} & = & \displaystyle{\sum_{n=0}^{\infty} \frac{(-2)^n}{e^n}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^{\infty} \bigg(\frac{-2}{e}\bigg)^n.}\\ \end{array}} |
| Step 2: |
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| Since 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 2<e,~\bigg|-\frac{2}{e}\bigg|<1.} |
| So, |
|
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=0}^{\infty} (-2)^ne^{-n}} & = & \displaystyle{\frac{1}{1+\frac{2}{e}}}\\ &&\\ & = & \displaystyle{\frac{1}{\frac{e+2}{e}}}\\ &&\\ & = & \displaystyle{\frac{e}{e+2}.}\\ \end{array}} |
(b)
| Step 1: |
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| This is a telescoping series. First, we find the partial sum of this series. |
| 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 s_k=\sum_{n=1}^k \bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg).} |
| Then, |
|
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=\frac{1}{2}-\frac{1}{2^{k+1}}.} |
| Step 2: |
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| Thus, |
|
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}\bigg(\frac{1}{2^n}-\frac{1}{2^{n+1}}\bigg)} & = & \displaystyle{\lim_{k\rightarrow \infty} s_k}\\ &&\\ & = & \displaystyle{\lim_{k\rightarrow \infty}\frac{1}{2}-\frac{1}{2^{k+1}}}\\ &&\\ & = & \displaystyle{\frac{1}{2}.}\\ \end{array}} |
| 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{e}{e+2}} |
| (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{1}{2}} |