Why does the series $\sum_{n=1}^\infty\frac1n$ not converge? Can someone give a simple explanation as to why the harmonic series 

$$\sum_{n=1}^\infty\frac1n=\frac 1 1 + \frac 12 + \frac 13 + \cdots $$

doesn't converge, on the other hand it grows very slowly? 
I'd prefer an easily comprehensible explanation rather than a rigorous proof regularly found in undergraduate textbooks.

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,
$$\begin{align}
\sum_{k=1}^\infty\frac1{k^s} & =\frac1{1-2^{1-s}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s} \\
\sum_{k=1}^\infty\frac1k & =\frac10\sum_{k=1}^\infty\frac{(-1)^{k+1}}k\tag{$s=1$} \\
& \to+\infty
\end{align}$$
