Using Euler's form of the Harmonic numbers,
$$\sum_{k=1}^n\frac1k=\int_0^1\frac{1-x^n}{1-x}dx$$
$$\begin{align} \lim_{n\to\infty}\sum_{k=1}^n\frac1k & =\lim_{n\to\infty}\int_0^1\frac{1-x^n}{1-x}dx \\ & =\int_0^1\frac1{1-x}dx \\ & =\left.\lim_{p\to1^+}-\ln(1-x)\right]_0^p \\ & \to+\infty \end{align}$$
Using the Taylor expansion of $\ln(1-x)$,
$$-\ln(1-x)=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\dots$$
$$-\ln(1-1)=1+\frac12+\frac13+\frac14+\dots\quad\ $$
Using Euler's relationship between the Riemann zeta function and the Dirichlet eta function,
$$\begin{align} \sum_{k=1}^\infty\frac1{k^s} & =\frac1{1-2^{1-s}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s} \\ \sum_{k=1}^\infty\frac1k & =\frac10\sum_{k=1}^\infty\frac{(-1)^{k+1}}k\tag{$s=1$} \\ & \to+\infty \end{align}$$