Difference between revisions of "009C Sample Final 2, Problem 4"
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<math>\begin{array}{rcl} | <math>\begin{array}{rcl} | ||
− | \displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x)^{n+1}}{(n+1)}\frac{n}{(-1)^n(x)^n}\bigg|}\\ | + | \displaystyle{\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|\frac{(-1)^{n+1}(x)^{n+1}}{(n+1)}\cdot\frac{n}{(-1)^n(x)^n}\bigg|}\\ |
&&\\ | &&\\ | ||
& = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n}{n+1}\bigg|}\\ | & = & \displaystyle{\lim_{n\rightarrow \infty} \bigg|(-1)(x)\frac{n}{n+1}\bigg|}\\ |
Revision as of 17:20, 10 March 2017
(a) Find the radius of convergence for the power series
(b) Find the interval of convergence of the above series.
Foundations: |
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Ratio Test |
Let be a series and |
Then, |
If the series is absolutely convergent. |
If the series is divergent. |
If the test is inconclusive. |
Solution:
(a)
Step 1: |
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We use the Ratio Test to determine the radius of convergence. |
We have |
|
Step 2: |
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The Ratio Test tells us this series is absolutely convergent if |
Hence, the Radius of Convergence of this series is |
(b)
Step 1: |
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First, note that corresponds to the interval |
To obtain the interval of convergence, we need to test the endpoints of this interval |
for convergence since the Ratio Test is inconclusive when |
Step 2: |
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First, let |
Then, the series becomes |
This is an alternating series. |
Let . |
The sequence is decreasing since |
for all |
Also, |
Therefore, this series converges by the Alternating Series Test |
and we include in our interval. |
Step 3: |
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Now, let |
Then, the series becomes |
This is a -series with Hence, the series diverges. |
Therefore, we do not include in our interval. |
Step 4: |
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The interval of convergence is |
Final Answer: |
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(a) The radius of convergence is |
(b) |