$\sum_{k=1}^{n} \frac{1}{k} \geq \int_1^{n+1} \frac{1}{x} dx$ Inequality $$\sum_{k=1}^{n} \frac{1}{k} \geq \int_1^{n+1} \frac{1}{x} dx$$
I don't see how you reach this inequality, or rather why it is correct. 
The context of this problem was the following:
Show that $$a_n := \sum_{k=1}^{n} \frac{1}{k} - \ln{n}$$ converges.
The prove that it is bounded, one can use the inequality above, to conclude that 
$$\sum_{k=1}^{n} \frac{1}{k} \geq \int_1^{n+1} \frac{1}{x} dx = \ln{(n+1)} \geq \ln{n}$$
 A: You can define a function $g(x)$ as
$$g(x) = \begin{cases}1 & 1\leq x< 2\\
\frac12 & 2\leq x < 3\\
\vdots & \vdots\\
\frac1n& n\leq x< n+1\end{cases}$$
which is a piecewise continuous (therefore integrable) function and $g(x)\geq \frac1x$ for all $x\in[1,n+1]$.
Because $g(x)\frac1x$, you also know that
$$\int_1^{n+1} g(x)dx \geq \int_1^{n+1} \frac1x dx$$
Now, the integral of $g$ can be split into $n$ integrals of $g$, i.e.
$$\int_1^{n+1} g(x)dx = \int_1^2 g(x)dx + \int_2^3g(x)dx + \dots + \int_n^{n+1} g(x) dx$$
and $g$ is constant on all these intervals.
A: By induction:


*

*$n=1$. $$\sum_{k=1}^1\frac1k=1>\int_{1}^{2}\frac1x dx=\ln 2\approx 0.69$$

*$n \to n+1$. Assume that the statement holds for $n\in \mathbb N$ and observe that $$\frac1{n+1}\int_{n+1}^{n+2}dx \ge \int_{n+1}^{n+2}\frac1xdx$$ since $\frac1x$ is decreasing on $[n+1,n+2]$. So \begin{align}\sum_{k=1}^{n+1}\frac1k=\frac{1}{n+1}+\sum_{k=1}^n\frac1k&\ge \frac{1}{n+1}\int_{n+1}^{n+2}dx+\int_{1}^{n+1}\frac1x dx \\[0.2cm]&\ge \int_{n+1}^{n+2}\frac1x dx+\int_{1}^{n+1}\frac1x dx=\int_{1}^{n+2}\frac1x dx\end{align}

