Limits problem (unable to solve further) Struggling to solve this problem,
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$
My approach:
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +...+ \frac{1}{6n}\right)$
= $\displaystyle\lim\limits_{n \to \infty} \int_{0}^{1}(x^{n-1} + x^{n} + x^{n-2}+...+ x^{6n-1}) dx$
= $\displaystyle\lim\limits_{n \to \infty}\int_{0}^{1}\left(x^{n-1} \cdot \frac{x^{5n+1} - 1}{x-1}\right)dx$
got stuck here and don't know how to solve it further (other approaches which are simpler would also help)

You also can use Riemann-sums:
$$
\left(\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}+\dots + \frac{1}{6n}\right)
$$
$$
=
\left(1+\frac{n}{n+1}+\frac{n}{n+2}+\dots + \frac{n}{n+ (5n-1)}\right)\frac{1}{n}+ \frac{1}{6n}
$$
$$
=\left(1+\frac{1}{1+1/n}+\frac{1}{1+2/n}+\dots + \frac{1}{1+(5n-1)/n}\right)\frac{1}{n} + \frac{1}{6n}
$$
$$
\to \int_0^5\frac{1}{1+x}dx + 0 = \ln(6) \quad (n \to \infty).
$$
Edit: On the interval $[0,5]$ we consider the partition
$$
x_k=\frac{k}{n} \quad (k=0, \dots, 5n),
$$
and the function $f(x)=1/(1+x)$. Choosing the left boundary point on each interval $[x_k,x_{k+1}]$ we get the Riemann-sum
$$
\sum_{k=0}^{5n-1} f(x_k) (x_{k+1}-x_k) = \sum_{k=0}^{5n-1} \frac{1}{1+k/n} \cdot \frac{1}{n}.
$$
Now it is known that
$$
\sum_{k=0}^{5n-1} f(x_k) (x_{k+1}-x_k) \to \int_0^5 f(x) dx \quad (n \to \infty);
$$
see https://en.wikipedia.org/wiki/Riemann_sum, for example.
