I have seen an interesting question on stackexchange, which I would like to requote so that I can understand the answer =)
$\lim\limits_{n\to\infty} \dfrac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}}$
How can the numerator be expressed as an integral?
$= \lim\limits_{n\to\infty} \dfrac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}}$
$=\lim\limits_{n\to\infty} \frac{\sum\limits_{k=1\to n} k^{99} }{n^{100}}$
$=\lim\limits_{n\to\infty} \frac{\sum\limits_{k=1\to n} k^{99} }{n^{100}}$
Any ideas?
Recall that if $f$ is integrable on $[a,b]$, then:
$$ \int_a^b f(x)~dx = \lim_{n\to \infty} \dfrac{b-a}{n}\sum_{k=1}^n f \left(a + k \left(\dfrac{b-a}{n}\right) \right) $$
Notice that: $$ \dfrac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}} = \sum_{k=1}^n \frac{k^{99}}{n^{100}} = \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^{99} = \frac{1 - 0}{n}\sum_{k=1}^n \left(0 + k\left(\frac{1 - 0}{n}\right)\right)^{99} $$ Hence, by taking $f(x) = x^{99}$ and $a = 0$ and $b = 1$, it follows that: $$ \lim\limits_{n\to\infty} \dfrac{1^{99} + 2^{99} + \cdots + n^{99}}{n^{100}} = \int_0^1 x^{99} \, dx = \left[ \frac{x^{100}}{100} \right]_0^1 = \frac{1}{100} $$
