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This question is more about the notation than the actual proof. My professor gave us the following proof:

$$\sum_{k=2}^{n} \frac{1}{k} \ge \sum_{k=2}^{n} \int_{k}^{k+1} \frac{1}{x} dx = \int_{2}^{n+1} \frac{1}{x} dx = ln(n+1)-ln(2) \rightarrow \infty $$

I struggle to understand what he did in the first 3 steps. I just don't see how the sum on the left is greater than the sum of the integrals, or why the limits of integration can simply be changed from k to 2 and from k+1 to n+1.

I'd appreciate any help!

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    $f(x):=\frac{1}{x}$ is positive and decreasing on $(0,\infty)$. So the integral from $k$ to $k+1$ is less than $f(k)\cdot\Delta_x=\frac{1}{k}\cdot((k+1)-k)=\frac{1}{k}$ (this is clear geometrically, you have a Riemann sum with only one rectangle). The sum equals the integral from $2$ to $n+1$ by linearity of the integral: you have the integral from $2$ to $3$ plus the integral from $3$ to $4$ plus eventually the integral from $n$ to $n+1$. – 2017-01-25

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The function $f(x)=\frac{1}{x}$ is continuous and decreasing on $\mathbb{R}^+$, hence for any $k\in\mathbb{N}^+$ we have $$ \frac{1}{k}>\int_{k}^{k+1}\frac{1}{x}\,dx \tag{1}$$ and by summing both sides of $(1)$ for $k=1,2,3,\ldots,n$ we get: $$ H_n = \sum_{k=1}^{n}\frac{1}{k} > \int_{1}^{n+1}\frac{dx}{x} = \log(n+1).\tag{2}$$ In the opposite direction, we may prove a slightly tighter inequality by exploiting $\frac{1}{k}<\log\left(\frac{2k+1}{2k-1}\right)$ and deducing $H_n < \log(2n+1)$.