# Thread: Evaluate The Following Limit. (2)

1. ## Evaluate The Following Limit. (2)

Hello Evaluate: $\displaystyle \lim_{n\to\infty} \frac{1}{n} [ (\frac{1}{n})^{9}+(\frac{2}{n})^{9} + (\frac{3}{n})^{9} + (\frac{4}{n})^{9} + .... + (\frac{n}{n})^{9} ]$

because I did not study it at my college.

2. ## Not sure

This is outside my area, but it looks like an application for L'Hopital's Rule.
Something to try unless someone else speaks up!

3. Originally Posted by Manx
This is outside my area, but it looks like an application for L'Hopital's Rule.
Something to try unless someone else speaks up!
it's a Riemann sum, plain and simple.

since the OP doesn't want to hear it, I'll tell you ...

the limit equals $\displaystyle \int_0^1 x^9 \, dx$

4. Originally Posted by TWiX
Hello Evaluate: $\displaystyle \lim_{n\to\infty} \frac{1}{n} [ (\frac{1}{n})^{9}+(\frac{2}{n})^{9} + (\frac{3}{n})^{9} + (\frac{4}{n})^{9} + .... + (\frac{n}{n})^{9} ]$

because I did not study it at my college.
I mean, friend. It's Riemann sums. Plain and simple. You can do a laborious argument to show that it's bounded above and below by terms that approach one-tenth.

Or, you can just note that this is $\displaystyle \lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\left(\f rac{k}{n}\right)^9=\lim_{n\to\infty}\frac{1}{n^{10 }}\sum_{k=1}^{n}k^9$. This sum can be proved by induction to be $\displaystyle \frac{1}{20} n^2 (n+1)^2 (n^2+n-1) (2 n^4+4 n^3-n^2-3 n+3)$. Take the limit and you will get one-tenth. Ta-da!

5. For the edification of the OP: Power Sum -- from Wolfram MathWorld