Partial sums of harmonic series

Question

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?

Answer 1

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}

Answer 2

Using Euler's form of the Harmonic numbers,

$$\sum_{k=1}^n\frac1k=\int_0^1\frac{1-x^n}{1-x}dx$$

$$\begin{align} \lim_{n\to\infty}\sum_{k=1}^n\frac1k & =\lim_{n\to\infty}\int_0^1\frac{1-x^n}{1-x}dx \\ & =\int_0^1\frac1{1-x}dx \\ & =\left.\lim_{p\to1^+}-\ln(1-x)\right]_0^p \\ & \to+\infty \end{align}$$

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Using the Taylor expansion of $\ln(1-x)$,

$$-\ln(1-x)=x+\frac{x^2}2+\frac{x^3}3+\frac{x^4}4+\dots$$

$$-\ln(1-1)=1+\frac12+\frac13+\frac14+\dots\quad\ $$

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Using Euler's relationship between the Riemann zeta function and the Dirichlet eta function,

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

Answer 3

There are also two additional methods.

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

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Method 2: Use Raabe–Duhamel's test (https://en.wikipedia.org/wiki/Convergence_tests)