Determine whether the following series converges or diverges.
| Foundations:
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| Review Ratio Test
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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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  .
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| First, we write the limit as Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {\lim _{n\rightarrow \infty }n\ln {\bigg (}{\frac {n}{n+1}}{\bigg )}}&=&\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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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