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anonymous

  • 5 years ago

i really need the answer.. how will i solve for the series n!/n^n to show it is converging/diverging

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  1. anonymous
    • 5 years ago
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    it converges

  2. anonymous
    • 5 years ago
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    use the ratio test

  3. anonymous
    • 5 years ago
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    \[\frac{(n+1)!}{(n+1)^{(n+1)}}\frac{n^n}{n!}\]=\[\frac{(n+1)n!}{(n+1)^n(n+1)}\frac{n^n}{n!}\]=\[\frac{n^n}{(n+1)^n}\]=\[\left(\frac{n}{n+1}\right)^n\]=\[(1-\frac{1}{n+1})^n\]

  4. anonymous
    • 5 years ago
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    now \[\lim_{n\rightarrow \infty}\left(1-\frac{1}{n+1}\right)^n\]

  5. anonymous
    • 5 years ago
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    i did this. we get 1 by this and it means that the test fails

  6. anonymous
    • 5 years ago
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    no you get \[frac{1}{e}\]

  7. anonymous
    • 5 years ago
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    \[\frac{1}{e}\]

  8. anonymous
    • 5 years ago
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    if you try and take the limit directly you get the indeterminate form \[1^\infty\] not 1

  9. anonymous
    • 5 years ago
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    you can evaluate the limit by capitalizing on the continuity of the natural logarithm function

  10. anonymous
    • 5 years ago
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    like so first set \[(1-\frac{1}{n+1})^n=L\] apply the natural log function to both sides to get \[ln (1-\frac{1}{n+1})^n=\ln L\]

  11. anonymous
    • 5 years ago
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    sorry I should be writing limit on the LHS

  12. anonymous
    • 5 years ago
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    even then the answer will be 1

  13. anonymous
    • 5 years ago
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    no its not

  14. anonymous
    • 5 years ago
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    how will you solve it?

  15. anonymous
    • 5 years ago
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    so we now use properties of logarithms

  16. anonymous
    • 5 years ago
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    \[\lim_{n\rightarrow\infty}n\ln(1-\frac{1}{n+1})=\ln L\] which still yields yet another indeterminate form \[\infty \times 0\] now rewrite to the equivalent expression \[\lim_{n\rightarrow\infty}\frac{\ln(1-\frac{1}{n+1})}{\frac{1}{n}}=\ln L\] so now we have indeterminate form \[\frac{0}{0}\] and now can finally use L'Hopital's rule

  17. anonymous
    • 5 years ago
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    apply L'Hopital's rule once we get \[\lim_{n\rightarrow\infty}\frac{\frac{1}{1-\frac{1}{n+1}}\frac{1}{(n+1)^2}}{-\frac{1}{n^2}}=\ln L\] simplifying we get \[\lim_{n\rightarrow\infty}\frac{-n^2}{(n+1)^2-(n+1)}=\ln L\] finally taking the limit we see that \[-1=\ln L\]

  18. anonymous
    • 5 years ago
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    exponentiating we see \[L=\frac{1}{e}\]

  19. anonymous
    • 5 years ago
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    that is a common result and easily verified with software

  20. anonymous
    • 5 years ago
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    ok.. thanks alot

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