Question

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 ...

Answer 1

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

Answer 2

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

Answer 3

well the last term looks like it will be one

Answer 4

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

Answer 5

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?

Answer 6

how did you get 1/10????

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

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

Answer 11

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

Answer 12

lol ,,,,

Answer 13

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

Answer 14

so how did u get 1/10???

Answer 15

got it

Answer 16

ur my hero !!!!

Answer 17

^^

Answer 18

can u help me with next one when i close LOL i got an answer. but idk if i got it right...pretty plzzz

Answer 19

i'll try. pls post.