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bahrom7893

Sequences and series

  • 2 years ago
  • 2 years ago

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  1. bahrom7893
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    I loathe them:

    • 2 years ago
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  2. bahrom7893
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    this one and one more left.

    • 2 years ago
  3. Zarkon
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    we know know that \[\int_{1}^{n}\frac{1}{x}dx=\ln(n)\] ...

    • 2 years ago
  4. Zarkon
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    |dw:1330845983664:dw| the sum of the rectangles is the sum of 1/x which is larger than the area under the curve which is the integral thus \[\sum_{k=1}^{n}\frac{1}{k}>\ln(n)\] ie.. \[\sum_{k=1}^{n}\frac{1}{k}-\ln(n)>0\]

    • 2 years ago
  5. bahrom7893
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    TY TY TY TY TY Zarkon.

    • 2 years ago
  6. Zarkon
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    \[\int_{n}^{n+1}\frac{1}{x}dx=\ln(n+1)-\ln(n)\] |dw:1330846312070:dw| the area of this rectangle is 1*(n+1)=n+1 and it is less than the area under the curve...thus... \[\frac{1}{n+1}<\ln(n+1)-\ln(n)\] \[\frac{1}{n+1}-\ln(n+1)<-\ln(n)\] \[\sum_{k=1}^{n}\frac{1}{n}+\frac{1}{n+1}-\ln(n+1)<\sum_{k=1}^{n}\frac{1}{n}-\ln(n)\] \[a_{n+1}<a_{n}\]

    • 2 years ago
  7. Zarkon
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    for part c since \(a_n>0\) and \(a_{n+1}<a_n\) by the monotone bounded sequence theorem \(a_n\) converges.

    • 2 years ago
  8. Zarkon
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    use you calculator or computer to get part (d)

    • 2 years ago
  9. bahrom7893
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    TY SOO MUCH! One more problem left for today.. I hope, unless i get stuck on computational ones.

    • 2 years ago
  10. Zarkon
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    it is getting really late here so it needs to be quick if you want my help

    • 2 years ago
  11. bahrom7893
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    ok sure one more and that's it then.

    • 2 years ago
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