Partial sums of harmonic series I've been given the following problem.

Prove that
$$\lim_{N\to\infty}\sum_{i=1}^N\frac1i=\infty.$$

In other words I need to prove that the partial sum of the harmonic series diverges.  I know the integral test works in this case, but does anybody know of any other methods for showing this?
 A: There are many ways to show this.
\begin{align}
& 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots \\[15pt]
= {} & 1 + \frac 1 3 + \frac 1 5 + \frac 1 7 + \cdots \\[8pt]
& \phantom{1} + \frac 1 2 + \frac 1 4 + \frac 1 6 + \cdots \\[15pt]
> {} & \phantom{{} + {}}  \frac 1 2 + \frac 1 4 + \frac 1 6 + \cdots \\[8pt]
& {} + \frac 1 2 + \frac 1 4 + \frac 1 6 + \cdots
\quad (\text{This “}{>}\text{'' is true if the sum is finite.)} \\[15pt]
= {} & 1 + \frac 1 2 + \frac 1 3 + \frac 1 4 + \frac 1 5 + \cdots
\end{align}
Here's a more frequently seen way:
\begin{align}
& 1 + \left(\frac 1 2\right) + \left(\frac 1 3 + \frac 1 4\right) + \left( \frac 1 5 + \cdots + \frac 1 8 \right) + \left( \frac 1 9 + \cdots + \frac 1 {16} \right) + \cdots \\[10pt]
\ge {} & 1 + \left(\frac 1 2\right) + \left(\frac 1 4 + \frac 1 4\right) + \left( \frac 1 8 + \cdots + \frac 1 8 \right) + \left( \frac 1 {16} + \cdots + \frac 1 {16} \right) + \cdots \\[10pt]
= {} & 1 + \frac 1 2 + \frac 1 2 + \frac 1 2 + \frac 1 2 + \cdots = \infty.  
\end{align}
A: There are also two additional methods.

Method 1: There is an necessary condition that states that a series of positive and decreasing $a_n$ can only be convergent if $\lim_{n\to \infty}n\cdot a_n=0$.

Method 2: Use Raabe–Duhamel's test (https://en.wikipedia.org/wiki/Convergence_tests)
A: 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,
$$\begin{align}
\sum_{k=1}^\infty\frac1{k^s} & =\frac1{1-2^{1-s}}\sum_{k=1}^\infty\frac{(-1)^{k+1}}{k^s} \\
\sum_{k=1}^\infty\frac1k & =\frac10\sum_{k=1}^\infty\frac{(-1)^{k+1}}k\tag{$s=1$} \\
& \to+\infty
\end{align}$$
