# Riemann sums to find limits

Lemma: If $F$ is continuous, then the sequence of arbitrary Riemann sums $$\sum \limits_{i=0}^{n-1} f(c_i)(x_{i+1}-x_i)$$ tends to $\int _a^b f$ as the norm of partition $||P|| \rightarrow 0$ for any choice of $c_i \in \Delta_i$.

Find $$\lim \limits_{n\rightarrow \infty} \frac1{n^6} \sum \limits_{k=0}^{n-1} k^5$$ Let $P_n$ be the partition of $[0,1]$ be $$\{0=\frac0n, \frac1n, \frac2n, \dots , \frac{n-1}n, \frac nn=1 \}$$ and $f_1(x)=x^5$.

By the lemma, $$\lim \limits_{n\rightarrow \infty} \frac1{n^6} \sum \limits_{k=0}^{n-1} k^5 = \lim \sum \limits_{k=0}^{n-1} \bigg(\frac{k}n \bigg)^5 \frac1n =\lim \sum \limits_{k=0}^{n-1} f_1(x_k)(x_{k+1}-x_k)= \int \limits_0^1 f_1$$

In the last expression, why does the second equality hold? And why did they choose the partition to be in $0$ and $1$?

Note that $x_k=k/n$ for the partition $P_n$ given, and that the subintervals all have the same width $x_{k+1}-x_k=(k+1)/n-k/n=1/n.$ Since the chosen function is $f_1(x)=x^5,$ this explains the second equality.
As for "why" they chose to partition $[0,1]$ it seems this is just an example of using the lemma in a specific case, and the given sum in the example fits into the lemma for the interval $[0,1]$ and equal partition, function $f_1(x)=x^5.$