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!
Answer
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 $H_n < \log(2n+1)$.
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