Difference between revisions of "009C Sample Midterm 2, Problem 5"
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!Step 2: | !Step 2: | ||
|- | |- | ||
− | | | + | |Let <math style="vertical-align: -5px">a\in (-2R,2R).</math> Then, <math style="vertical-align: -13px">\frac{a}{2} \in (-R,R).</math> |
+ | |- | ||
+ | |So, | ||
+ | |- | ||
+ | | <math>\sum_{n=0}^\infty c_n\bigg(\frac{a}{2}\bigg)^n</math> | ||
+ | |- | ||
+ | |converges by assumption. | ||
+ | |- | ||
+ | |Since <math style="vertical-align: 0px">a</math> was an arbitrary number in the interval <math style="vertical-align: -5px">a\in (-2R,2R),</math> | ||
+ | |- | ||
+ | | <math>\sum_{n=0}^\infty c_n\bigg(\frac{x}{2}\bigg)^n</math> | ||
+ | |- | ||
+ | |converges in the interval <math style="vertical-align: -5px">(-2R,2R).</math> | ||
|} | |} | ||
Revision as of 16:19, 23 April 2017
If converges, does it follow that the following series converges?
(a)
(b)
Foundations: |
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If a power series converges, then it has a nonempty interval of convergence. |
Solution:
(a)
Step 1: |
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Assume that the power series converges. |
Let be the radius of convergence of this power series. |
So, the power series |
converges in the interval |
Step 2: |
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Let Then, |
So, |
converges by assumption. |
Since was an arbitrary number in the interval |
converges in the interval |
(b)
Step 1: |
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Step 2: |
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Final Answer: |
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(a) converges |
(b) converges |