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)
 A: 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.
A: I will use the fact that the sequence
$$x_n = 1+\frac{1}{2}+\ldots+\frac{1}{n}-\ln n$$
has a limit $\gamma = 0.577...$ (known as Euler–Mascheroni constant). Let's denote
$$a_n = \frac{1}{n}+\ldots+\frac{1}{6n}.$$
We can see that
$$x_{6n}-x_{n-1}=a_n-\ln(6n)+\ln(n-1) = a_n -\ln\left(\frac{6n}{n-1}\right).$$
Then
$$\lim_{n\to\infty}a_n = \lim_{n\to\infty}\left(x_{6n}-x_{n-1}+\ln\left(\frac{6n}{n-1}\right)\right) = \gamma-\gamma+\ln6 = \ln 6.$$
A: 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:
$\displaystyle \lim\limits_{n \to \infty} \left(\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} +\dots+ \frac{1}{6n}\right)$=
$\displaystyle \lim\limits_{n \to \infty} H_{6n}-H_n$ =
$\lim\limits_{n \to \infty}$ ($ln(6n)+\gamma+\frac{1}{6n}-\epsilon_{6n}-(ln(n)+\gamma+\frac{1}{2n}-\epsilon_n)$)=
$\lim\limits_{n \to \infty}$ ($ln(6)+ln(n)+\gamma+\frac{1}{6n}-\epsilon_{6n}-ln(n)-\gamma-\frac{1}{2n}+\epsilon_n$)= $ln(6)$.
A: This method uses neither the Euler-Mascheroni constant, nor Riemann sums. Rather, it uses bounds from the graph of $\ f(x) = \frac{1}{x},\ $ using the fact that $\ f\ $ is decreasing towards $\ 0.$
Whenever $\ k\geq i\geq 1,\ $ we have:
$$\int_{x=i+1}^{x=k+1} \frac{1}{x} dx \leq \sum_{j=i+1}^{j=k} \frac{1}{j} \leq \int_{x=i}^{x=k} \frac{1}{x} dx $$
$$ \implies \ln(k+1) - \ln(i+1) \leq \sum_{j=i+1}^{j=k} \frac{1}{j} \leq \ln(k) - \ln(i). \qquad (1) $$
If $\ n\ $ is large enough, we can substitute $\ k=6n\ $ and $\ k=n-1\ $ into $\ (1)\ $ giving, respectively:
$$ \ln(6n+1) - \ln(i+1) \leq \sum_{j=i+1}^{j=6n} \frac{1}{j} \leq \ln(6n) - \ln(i) \qquad (2) $$
and
$$ \ln(n) - \ln(i+1) \leq \sum_{j=i+1}^{j=n-1} \frac{1}{j} \leq \ln(n-1) - \ln(i). \qquad (3) $$
$(2),\ (3)\implies$
$$\ln\left(\frac{6n+1}{n-1}\right) + \ln\left(\frac{i}{i+1}\right) \leq \sum_{j=n}^{j=6n} \frac{1}{j} \leq \ln\left(\frac{6n}{n}\right) + \ln\left(\frac{i+1}{i}\right). \qquad (4) $$
Taking limits in $\ (4),\ $ first as $\ n\to\infty,\ $ then as $\ i\to\infty,\ $ we get:
$$\ln 6 \leq \lim_{n\to\infty}\sum_{j=n}^{j=6n} \frac{1}{j} \leq \ln6. $$
