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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?

Community
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George Wilson
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  • Partial sum or infinite sum? – imranfat Oct 06 '16 at 15:53
  • Search the website for "Harmonic." There are many proofs that $H_n \sim \ln n$, and since $\ln n \to \infty$... – Clement C. Oct 06 '16 at 15:55
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    https://math.stackexchange.com/questions/215907/another-proof-that-harmonic-series-diverges, https://math.stackexchange.com/questions/1130923/proof-that-the-harmonic-series-diverges-without-improper-integrals, https://math.stackexchange.com/questions/1501287/alternative-ways-to-show-that-the-harmonic-series-diverges, https://math.stackexchange.com/questions/180903/my-proof-that-a-harmonic-series-diverges, https://math.stackexchange.com/questions/1130923/proof-that-the-harmonic-series-diverges-without-improper-integrals, you get the point... – Eman Yalpsid Oct 06 '16 at 15:56
  • It's too easy to look up $n$ kinds of proofs in $N$ number of texts in a library if you willing to do so. – Ng Chung Tak Oct 06 '16 at 17:35

3 Answers3

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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}

Michael Hardy
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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)

MrYouMath
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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,

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

Simply Beautiful Art
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