Determine whether the following series converges or diverges.
- 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{n!}{n^n}}
| Foundations:
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| Recall:
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| 1. Ratio Test 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 \sum a_n}
be a series and 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 L=\lim_{n\rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|}
. Then,
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- If 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 L<1}
, the series is absolutely convergent.
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- If 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 L>1}
, the series is divergent.
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- If
 , the test is inconclusive.
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| 2. If a series absolutely converges, then it also converges.
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Solution:
| Step 1:
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| We proceed using the ratio test.
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| 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}(n+1)!}{(n+1)^{n+1}}\frac{n^n}{(-1)^n n!}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)n!}{n!}\frac{n^n}{(n+1)^{n+1}}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\frac{(n+1)n^n}{(n+1)(n+1)^n}\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg|\bigg(\frac{n}{n+1}\bigg)^n\bigg|}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty}\bigg(\frac{n}{n+1}\bigg)^n.}\\ \end{array}}
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| Step 2:
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| Now, we continue to calculate the limit from Step 1. 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{n}{n+1}\bigg)^n}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty}e^{\ln(\frac{n}{n+1})^n}}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty}e^{n\ln(\frac{n}{n+1})}}\\ &&\\ & = & \displaystyle{e^{\lim_{n \rightarrow \infty}n\ln(\frac{n}{n+1})}.}\\ \end{array}}
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| Step 3:
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| Now, we need to 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_{n \rightarrow \infty}n\ln\bigg(\frac{n}{n+1}\bigg).}
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| First, we write the limit as 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_{n \rightarrow \infty}\frac{\ln\bigg(\frac{n}{n+1}\bigg)}{\frac{1}{n}}.}
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| Now, we use L'Hopital's Rule to get
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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}n\ln\bigg(\frac{n}{n+1}\bigg)} & \overset{l'H}{=} & \displaystyle{\lim_{n \rightarrow \infty}\frac{\frac{n+1}{n}\frac{(n+1)-n}{(n+1)^2}}{-\frac{1}{n^2}}}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty} \frac{1}{n(n+1)}(-n^2)}\\ &&\\ & = & \displaystyle{\lim_{n \rightarrow \infty} \frac{-n}{n+1}}\\ &&\\ & = & \displaystyle{-1.}\\ \end{array}}
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| Step 4:
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| We go back to Step 2 and use the limit we calculated in Step 3.
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| 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 \lim_{n \rightarrow \infty}\bigg|\frac{a_{n+1}}{a_n}\bigg|=e^{-1}=\frac{1}{e}<1.}
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| Thus, the series absolutely converges by the Ratio Test.
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| Since the series absolutely converges, the series also converges.
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| Final Answer:
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| The series converges.
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