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10.5 Harmonic Series and p-Series

Keywords

Harmonic Series 调和级数 • p-Series p-级数

Riemann Zeta Function 黎曼Zeta函数

The Harmonic Series

The Harmonic Series is the sum of the reciprocals of the positive integers:

\[ \sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots \]

Even though the terms \(\frac{1}{n}\) approach zero, the series diverges. It grows toward infinity, albeit very slowly.

Proof: Why the Harmonic Series Diverges

We can prove divergence using a "grouping" method (originally discovered by Nicole Oresme). We compare the Harmonic series to a series that we know diverges because it adds up infinitely many halves.

  1. Write out the series: \(S = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \dots\)

  2. Group the terms into powers of 2: \(S = 1 + \left( \frac{1}{2} \right) + \left( \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) + \dots\)

  3. Replace terms with smaller values:

    • In the second group: \(\frac{1}{3} + \frac{1}{4} > \frac{1}{4} + \frac{1}{4} = \frac{1}{2}\)
    • In the third group: \(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} > \frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{1}{2}\)

  4. Establish the inequality: \(S > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots\)

Since we are adding \(1/2\) infinitely many times, the sum must be infinite. Therefore, the Harmonic series diverges.

The p-Series Test

The Harmonic series is a specific case (\(p=1\)) of a broader family called \(p\)-series.

核心要点：The p-Series Test (LIM-7.A.7)

A series of the form \(\sum\limits_{n=1}^{\infty} \frac{1}{n^p}\) is called a \(p\)-series, where \(p\) is a positive constant.

  • Converges if \(p > 1\).
  • Diverges if \(p \le 1\).

Notice that \(1/n^2\) (blue) approaches the x-axis much faster than \(1/n\) (orange). That extra speed is what allows the sum to converge to a finite number.

Extension：The Basel Problem

In Section 1.8, we used the Squeeze Theorem to prove \(\lim\limits_{x \to 0} \frac{\sin x}{x} = 1\). In Section 10.4, we used the Integral Test to prove that the \(p\)-series \(\sum\limits_{n=1}^{\infty} \frac{1}{n^2}\) converges.

But exactly what does it converge to? This question baffled the greatest mathematicians for 90 years until 1734, when a 28-year-old Leonhard Euler provided a solution that shocked the mathematical world.

1.The Challenge

Find the exact sum of the reciprocals of the squares of all positive integers:

\[ S = 1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \dots = \sum_{n=1}^{\infty} \frac{1}{n^2} \]

2.Euler’s Intuition (The Leap of Faith)

Euler solved this by comparing two different ways to write the function \(\frac{\sin x}{x}\):

  • The Taylor Series Representation:

    From Unit 10, we know the expansion of \(\sin x\):

    \[ \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \dots \]

    Dividing by \(x\), we get:

    \[ \frac{\sin x}{x} = 1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \dots \]

  • The Infinite Product Representation:

    Euler noticed that \(\frac{\sin x}{x} = 0\) precisely when \(x = \pm \pi, \pm 2\pi, \pm 3\pi, \dots\). Analogous to how a polynomial \(P(x)\) with roots \(r_1, r_2\) can be factored as \((1 - x/r_1)(1 - x/r_2)\), he boldly factored the infinite-degree "polynomial" of \(\sin x / x\) using its roots:

    \[ \frac{\sin x}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \dots \]

3.Comparing the Coefficients If you were to expand this infinite product and look only at the \(x^2\) term, you would get the sum of all the individual \(x^2\) parts:

\[ \text{Coefficient of } x^2 = -\left( \frac{1}{\pi^2} + \frac{1}{4\pi^2} + \frac{1}{9\pi^2} + \dots \right) = -\frac{1}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2} \]

From the Taylor series above, we already know the coefficient of \(x^2\) is \(-1/3!\) (which is \(-1/6\)).

Setting these two coefficients equal:

\[ -\frac{1}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2} = -\frac{1}{6} \]

4.The Final Result

\[ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6} \]

This result was mind-blowing because it introduced \(\pi\)—a constant from circles and geometry—into a problem involving pure integers. This not only solved the Basel Problem but also laid the foundation for the Riemann Zeta Function, specifically \(\zeta(2) = \frac{\pi^2}{6}\).

Want to see a visual proof? Check out 3Blue1Brown’s geometric solution, which explains this result using the inverse-square law of light without relying on complex series expansions.

Extension: The Riemann Zeta Function & The Prime Mystery

If we replace the real number \(p\) with a complex variable \(s\), we get the Riemann Zeta Function:

\[ \zeta(s) = \sum_{n=1}^{\infty} \frac{1}{n^s} = 1 + \frac{1}{2^s} + \frac{1}{3^s} + \dots \]

In 1859, Bernhard Riemann discovered a deep connection between the zeros of this function and the distribution of prime numbers.

The Riemann Hypothesis conjectures that all "non-trivial" zeros of this function lie on the "critical line" where the real part of \(s\) is \(1/2\). This is one of the seven Millennium Prize Problems; proving it would earn you $1,000,000 and revolutionize number theory.

Learn more about it at the Clay Mathematics Institute.

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