## jackkim4 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 ... one year ago one year ago

1. UnkleRhaukus

$\lim\limits_{n\rightarrow\infty} \frac1n\left[\left(\frac1n\right)^9+\left(\frac2n\right)^9+\left(\frac3n\right)^9+...+\left(\frac nn\right)^9\right]$

2. jackkim4

yes! it's not 0 is it...

3. UnkleRhaukus

well the last term looks like it will be one

4. sirm3d

1/10 is my answer. is it among the choices?

5. jackkim4

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

how did you get 1/10????

7. sirm3d

$\large \int\limits_{a}^{b}f(x) dx=\lim_{n \rightarrow +\infty}\sum_{i=1}^{n} f(a+i \Delta x)\Delta x,\quad \Delta x = \frac{ b-a }{ n }$

8. jackkim4

i has no idea wat the triangle stands for :'(

9. sirm3d

$\left( \frac{ 1 }{ n } \right)^9+...+\left( \frac{ n }{ n } \right)^9=\sum_{i=1}^{n}\left( \frac{ i }{ n } \right)^n$

10. jackkim4

i think the way ur doing it is correct... ur using the rem. sum right?

11. sirm3d

$\large \sum_{i=1}^{n}\left( \frac{ i }{ n } \right)^9=\sum_{i=1}^{n}\left( 0+i\frac{ 1 }{ n } \right)^9$so a = 0, b=1 (delta x) = 1/n

12. jackkim4

lol ,,,,

13. sirm3d

therefore $\large f(x)=x^9$\ YES, the problem is a riemann sum.

14. jackkim4

so how did u get 1/10???

15. jackkim4

got it

16. jackkim4

ur my hero !!!!

17. sirm3d

^^

18. jackkim4

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

i'll try. pls post.