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nethravathy

show that the sequence xn=1/n+1+1/n+2+---+1/n+n converge

  • 11 months ago
  • 11 months ago

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  1. nethravathy
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    pls solve this problem

    • 11 months ago
  2. perl
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    |dw:1366719299230:dw|

    • 11 months ago
  3. nethravathy
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    yes that is the problem

    • 11 months ago
  4. joemath314159
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    You can rewrite the series as:\[\frac{\frac{1}{n}}{1+\frac{1}{n}}+\frac{\frac{1}{n}}{1+\frac{2}{n}}+\frac{\frac{1}{n}}{1+\frac{3}{n}}+\cdots+\frac{\frac{1}{n}}{1+\frac{n}{n}}\]\[=\frac{1}{1+\frac{1}{n}}\cdot \frac{1}{n}+\frac{1}{1+\frac{2}{n}}\cdot \frac{1}{n}+\cdots +\frac{1}{1+\frac{n}{n}}\cdot \frac{1}{n}\]Then it should be clear that this series is actually a Riemann sum. Its getting the areas under the function:\[f(x)=\frac{1}{1+x}\] over the interval [0,1] by splitting the interval into n equal pieces. So the limit as n goes to infinity is:\[\int\limits_{0}^1\frac{1}{1+x}dx\]

    • 11 months ago
  5. nethravathy
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    thank you next question

    • 11 months ago
  6. nethravathy
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    expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion

    • 11 months ago
  7. perl
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    @joemath314159 hello, how didyou get that answer

    • 11 months ago
  8. perl
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    |dw:1366720645769:dw|

    • 11 months ago
  9. perl
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    thats as far as i got

    • 11 months ago
  10. nethravathy
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    is this the answer

    • 11 months ago
  11. perl
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    joe got it, i was wondering how

    • 11 months ago
  12. perl
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    |dw:1366720894172:dw|

    • 11 months ago
  13. perl
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    not only does it converge but we can find what it converges to

    • 11 months ago
  14. nethravathy
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    ln2 is the answer or what

    • 11 months ago
  15. perl
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    it only asked if it converged . and yes it does

    • 11 months ago
  16. nethravathy
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    joe is correct or yours

    • 11 months ago
  17. joemath314159
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    You want to look at it as:\[x_n=\sum_{k=1}^{n}\frac{1}{n+k}=\sum_{k=1}^{n}\frac{\frac{1}{n}}{1+\frac{k}{n}}=\sum_{k=1}^{n}\frac{1}{1+\frac{k}{n}}\cdot\frac{1}{n}\]Then let:\[f(x)=\frac{1}{1+x}\]Using this, we can change the sum to:\[\sum_{k=1}^{n}f\left(\frac{k}{n}\right)\cdot\left(\frac{k}{n}-\frac{k-1}{n}\right)\].This is a Riemann sum of f(x) over the interval [0,1]. If you split the interval into n equal pieces, each piece is 1/n long. so thats the width of a rectangle. The height is the value of the functions at the point x = k/n.

    • 11 months ago
  18. joemath314159
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    Skip to 4:30 of this video, it should explain the idea with a similar problem. http://www.youtube.com/watch?v=xt86sFvvdhA

    • 11 months ago
  19. perl
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    joemath, i see that. im just not sure how you came to this solution.

    • 11 months ago
  20. perl
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    i will check it, thanks

    • 11 months ago
  21. perl
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    and is there another way to solve this?

    • 11 months ago
  22. joemath314159
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    im not sure if there is another way to get the actual limit. This might be the only way to get it by hand.

    • 11 months ago
  23. perl
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    |dw:1366721812063:dw|

    • 11 months ago
  24. joemath314159
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    Thats right. Let a = 0 and b = 1.

    • 11 months ago
  25. perl
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    |dw:1366721876414:dw|

    • 11 months ago
  26. perl
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    well you are a genius to have solved this. this is not obvious to use this approach. any luck with the second question the maclaurin?

    • 11 months ago
  27. perl
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    expand xy2+cos[xy] upto fourth degree terms using maclaurin's expansion

    • 11 months ago
  28. nethravathy
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    ya thats the second question

    • 11 months ago
  29. perl
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    we know that cos x = 1 - x^2/2!+ x^4/4!+...

    • 11 months ago
  30. joemath314159
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    I haven't been thinking about it =/ I'm getting ready to head to class >.<

    • 11 months ago
  31. perl
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    cya :)

    • 11 months ago
  32. perl
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    @nethravathy |dw:1366722117226:dw|

    • 11 months ago
  33. nethravathy
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    ya the same

    • 11 months ago
  34. perl
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    |dw:1366722262487:dw|

    • 11 months ago
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