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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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 ...

Mathematics
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\[\lim\limits_{n\rightarrow\infty} \frac1n\left[\left(\frac1n\right)^9+\left(\frac2n\right)^9+\left(\frac3n\right)^9+...+\left(\frac nn\right)^9\right] \]
yes! it's not 0 is it...
well the last term looks like it will be one

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Other answers:

1/10 is my answer. is it among the choices?
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?
how did you get 1/10????
\[\large \int\limits_{a}^{b}f(x) dx=\lim_{n \rightarrow +\infty}\sum_{i=1}^{n} f(a+i \Delta x)\Delta x,\quad \Delta x = \frac{ b-a }{ n }\]
i has no idea wat the triangle stands for :'(
\[\left( \frac{ 1 }{ n } \right)^9+...+\left( \frac{ n }{ n } \right)^9=\sum_{i=1}^{n}\left( \frac{ i }{ n } \right)^n\]
i think the way ur doing it is correct... ur using the rem. sum right?
\[\large \sum_{i=1}^{n}\left( \frac{ i }{ n } \right)^9=\sum_{i=1}^{n}\left( 0+i\frac{ 1 }{ n } \right)^9\]so a = 0, b=1 (delta x) = 1/n
lol ,,,,
therefore \[\large f(x)=x^9\]\ YES, the problem is a riemann sum.
so how did u get 1/10???
got it
ur my hero !!!!
^^
can u help me with next one when i close LOL i got an answer. but idk if i got it right...pretty plzzz
i'll try. pls post.

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