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)

Let define $\displaystyle H_n=1+\frac {1}{2}+ \frac {1}{3}...+\frac {1}{n}
 $ more precisely is the sum up to n $H_n=\sum_{k=0}^n=\frac {1}{k}$.
Using Euler-Maclaurin formula $\displaystyle H_n=ln(n)+\gamma+\frac{1}{2n}-\epsilon_n$ where $\gamma\approx0.5772...$and $0\le\epsilon_n\le\frac{1}{8n^2}$
$\displaystyle H_{6n}=ln(6n)+\gamma+\frac{1}{6n}-\epsilon_{6n}$
Noting that your expression is equal to:
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$=
$\displaystyle \lim\limits_{n \to \infty} H_{6n}-H_n$ =
$\lim\limits_{n \to \infty}$ ($ln(6n)+\gamma+\frac{1}{6n}-\epsilon_{6n}-(ln(n)+\gamma+\frac{1}{2n}-\epsilon_n)$)=
$\lim\limits_{n \to \infty}$ ($ln(6)+ln(n)+\gamma+\frac{1}{6n}-\epsilon_{6n}-ln(n)-\gamma-\frac{1}{2n}+\epsilon_n$)= $ln(6)$.
