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jackkim4

  • 2 years ago

Plz help! super challenging question... Evaluate: lim n->inf 1/n[(1/n)^9+(2/n)^9+(3/n)^9+...+(n/n)^9] p.s: i got 0...but i don't think that's right ...

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  1. UnkleRhaukus
    • 2 years ago
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    \[\lim\limits_{n\rightarrow\infty} \frac1n\left[\left(\frac1n\right)^9+\left(\frac2n\right)^9+\left(\frac3n\right)^9+...+\left(\frac nn\right)^9\right] \]

  2. jackkim4
    • 2 years ago
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    yes! it's not 0 is it...

  3. UnkleRhaukus
    • 2 years ago
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    well the last term looks like it will be one

  4. sirm3d
    • 2 years ago
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    1/10 is my answer. is it among the choices?

  5. jackkim4
    • 2 years ago
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    i separated (1)^9/(n)^9 and factored out all the (n)^9. resulting in (1/n^10) in the front. so that (1/n^10) times (n)^9 the last term will cancel. which will b left with (1/n) and (1/inf) is 0. 0 times anything left in the [ ...] will be 0. right?

  6. jackkim4
    • 2 years ago
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    how did you get 1/10????

  7. sirm3d
    • 2 years ago
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    \[\large \int\limits_{a}^{b}f(x) dx=\lim_{n \rightarrow +\infty}\sum_{i=1}^{n} f(a+i \Delta x)\Delta x,\quad \Delta x = \frac{ b-a }{ n }\]

  8. jackkim4
    • 2 years ago
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    i has no idea wat the triangle stands for :'(

  9. sirm3d
    • 2 years ago
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    \[\left( \frac{ 1 }{ n } \right)^9+...+\left( \frac{ n }{ n } \right)^9=\sum_{i=1}^{n}\left( \frac{ i }{ n } \right)^n\]

  10. jackkim4
    • 2 years ago
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    i think the way ur doing it is correct... ur using the rem. sum right?

  11. sirm3d
    • 2 years ago
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    \[\large \sum_{i=1}^{n}\left( \frac{ i }{ n } \right)^9=\sum_{i=1}^{n}\left( 0+i\frac{ 1 }{ n } \right)^9\]so a = 0, b=1 (delta x) = 1/n

  12. jackkim4
    • 2 years ago
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    lol ,,,,

  13. sirm3d
    • 2 years ago
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    therefore \[\large f(x)=x^9\]\ YES, the problem is a riemann sum.

  14. jackkim4
    • 2 years ago
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    so how did u get 1/10???

  15. jackkim4
    • 2 years ago
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    got it

  16. jackkim4
    • 2 years ago
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    ur my hero !!!!

  17. sirm3d
    • 2 years ago
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    ^^

  18. jackkim4
    • 2 years ago
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    can u help me with next one when i close LOL i got an answer. but idk if i got it right...pretty plzzz

  19. sirm3d
    • 2 years ago
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    i'll try. pls post.

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