<s>[INST] Let $p_n$ be the sequence defined by $p_n=\sum_{k=1}^n\frac{1}{k}$. Show that $p_n$ diverges even though $\lim_{n\to\infty}(p_n-p_{n-1})=0$ I have tried this as :
$$p_n=\sum_{k=1}^n\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}+\frac{1}{n}$$
$$p_{n-1}=\sum_{k=1}^{n-1}\frac{1}{k}=1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{n-1}$$
$$(p_n-p_{n-1})=\frac{1}{n}$$
$$\lim_{n\to\infty}(p_n-p_{n-1})=\lim_{n\to\infty}\frac{1}{n}=0$$
But dont know how to show $p_n$ diverges?
 [/INST] $$\frac{1}{3}+\frac{1}{4}>\frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$
Generalising, $$\frac{1}{2^n-2^{n-1}+1}+\dots +\frac{1}{2^n}>\frac{1}{2^n}+\dots +\frac{1}{2^n}\text{ ($2^{n-1}$ terms)}$$
Hence, we can add an infinite number of $\frac{1}{2}$. The sum thus diverges. 
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