Hello Evaluate: $\displaystyle \lim_{n\to\infty} \frac{1}{n} [ (\frac{1}{n})^{9}+(\frac{2}{n})^{9} + (\frac{3}{n})^{9} + (\frac{4}{n})^{9} + .... + (\frac{n}{n})^{9} ]$
Please do not talk about Raiman Sums.
because I did not study it at my college.
Hello Evaluate: $\displaystyle \lim_{n\to\infty} \frac{1}{n} [ (\frac{1}{n})^{9}+(\frac{2}{n})^{9} + (\frac{3}{n})^{9} + (\frac{4}{n})^{9} + .... + (\frac{n}{n})^{9} ]$
Please do not talk about Raiman Sums.
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!
For the edification of the OP: Power Sum -- from Wolfram MathWorld