Find the interval of convergence of the following series.
| Foundations:
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| Ratio Test
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| Check endpoints of interval
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Solution:
| Step 1:
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| We proceed using the ratio test to find the interval of convergence. So, we have
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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{\lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|} & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(-1)^{n+1}(x+2)^{n+1}}{(n+1)^2}}\frac{n^2}{(-1)^n(x+2)^n}\bigg|\\ &&\\ & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\frac{n^2}{(n+1)^2}}\\ &&\\ & = & \displaystyle{|x+2|\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^2}\\ &&\\ & = & \displaystyle{|x+2|\bigg(\lim_{n \rightarrow \infty}\frac{n}{n+1}\bigg)^2}\\ &&\\ & = & \displaystyle{|x+2|(1)^2}\\ &&\\ & = & \displaystyle{|x+2|}\\ \end{array}}
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| Step 2:
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| 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 |x+2|<1}
. Hence, our interval 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 (-3,-1)}
. But, we still need to check the endpoints of this interval
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| to see if they are included in the interval of convergence.
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| Step 3:
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| First, 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=-1}
. Then, our series becomes 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} (-1)^n \frac{1}{n^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 n^2<(n+1)^2}
, 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 \frac{1}{(n+1)^2}<\frac{1}{n^2}}
. 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 \frac{1}{n^2}}
is decreasing.
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| 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 sum_{n=0}^{\infty} (-1)^n \frac{1}{n^2}}
converges by the Alternating Series Test.
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| Step 4:
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| Now, 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=-3}
. Then, our series becomes
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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{\sum_{n=0}^{\infty} (-1)^n \frac{(-1)^n}{n^2}} & = & \displaystyle{\sum_{n=0}^{\infty} (-1)^{2n} \frac{1}{n^2}}\\ &&\\ & = & \displaystyle{\sum_{n=0}^{\infty} \frac{1}{n^2}}\\ \end{array}}
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| This is a convergent series by the p-test.
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| Step 5:
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| Thus, the interval of convergence for this 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 [-3,-1]}
.
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| Final Answer:
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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 [-3,-1]}
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