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?

Recall the inequality $e^x>1+x \forall x>0$
Now Consider $$e^{p_n}=e^{\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)}$$ $$=e\cdot e^{\frac{1}{2}}\cdot e^{\frac{1}{3}}\cdots e^{\frac{1}{n}}$$ $$>\left(1+1\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)\cdots\left(1+\frac{1}{n}\right)$$ $$=(2)\left(\frac{3}{2}\right)\left(\frac{4}{3}\right)\cdots\left(\frac{n+1}{n}\right)$$ $$=n+1$$
Thus, taking $\ln$ on both sides (inequality maintains as $\ln$ is an increasing function),
$$p_n>\ln(n+1)$$ and $\ln(n+1)$ is unbounded as $n\uparrow$ and hence $p_n$ diverges.
