Difference between revisions of "009C Sample Final 2, Problem 2"
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<span class="exam"> For each of the following series, find the sum if it converges. If it diverges, explain why. | <span class="exam"> For each of the following series, find the sum if it converges. If it diverges, explain why. | ||
| − | <span class="exam">(a) <math>4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math> | + | <span class="exam">(a) <math style="vertical-align: -14px">4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\cdots</math> |
<span class="exam">(b) <math>\sum_{n=1}^{+\infty} \frac{1}{(2n-1)(2n+1)}</math> | <span class="exam">(b) <math>\sum_{n=1}^{+\infty} \frac{1}{(2n-1)(2n+1)}</math> | ||
Revision as of 18:13, 4 March 2017
For each of the following series, find the sum if it converges. If 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 4-2+1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\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{1}{(2n-1)(2n+1)}}
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Solution:
(a)
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(b)
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| (a) |
| (b) |