Harmonic Series and p-Series

The harmonic series $1 + \tfrac{1}{2} + \tfrac{1}{3} + \tfrac{1}{4} + \dots$ is one of the most surprising results in math: its terms shrink to $0$, yet the sum is infinite. It belongs to a whole family, the p-series $\sum \frac{1}{n^p}$, and there’s one simple rule for which members converge. These series are the measuring sticks you’ll compare other series against in the comparison tests, so it pays to know them cold.

Key ideas

The p-series rule

A p-series has the form

$$\sum_{n=1}^{\infty} \frac{1}{n^p} = 1 + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \dots$$

where $p$ is a constant. Then:

Value of $p$	The p-series
$p \gt 1$	converges
$0 \lt p \le 1$	diverges (terms go to $0$, but too slowly)
$p \le 0$	diverges by the $n$th term test (terms don’t go to $0$)

Why: the integral test

For $p \gt 0$, $f(x) = \dfrac{1}{x^p}$ is positive, continuous, and decreasing for $x \ge 1$, so the integral test applies. And you know which of these improper integrals converge:

$$\int_1^{\infty} \frac{1}{x^p}\,dx = \begin{cases} \dfrac{1}{p - 1} & \text{if } p \gt 1 \\[2ex] \infty & \text{if } p \le 1 \end{cases}$$

(Remember, the value $\frac{1}{p-1}$ is the integral, not the sum of the series.)

The harmonic series

When $p = 1$ you get the harmonic series $\sum \frac{1}{n}$, which diverges. It just barely diverges, though. The partial sums grow about as fast as $\ln n$, so it takes $12\,367$ terms for the sum to pass $10$.

Partial sums of $\sum \frac{1}{n}$ keep growing (slowly), while those of $\sum \frac{1}{n^2}$ level off at $\frac{\pi^2}{6}$.

Recognizing a p-series in disguise

Rewrite roots and products as a single power of $n$:

$$\frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}}, \qquad \frac{\sqrt{n}}{n^2} = \frac{1}{n^{3/2}}, \qquad \frac{1}{n\sqrt[3]{n}} = \frac{1}{n^{4/3}} .$$

A constant multiple doesn’t change convergence, so $\sum \frac{5}{n^3}$ converges just like $\sum \frac{1}{n^3}$.

Don’t confuse p-series with geometric series. In a p-series the variable is in the base ($\frac{1}{n^2}$); in a geometric series it’s in the exponent ($\frac{1}{2^n}$).

Worked examples

Example 1: Using the rule

Does each series converge or diverge?

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^4}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt{n}}$
• (c) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{1.01}}$
• (d) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2\sqrt{n}}$

Solution.

(a) $p = 4 \gt 1$: converges.

(b) $\dfrac{1}{\sqrt{n}} = \dfrac{1}{n^{1/2}}$, so $p = \tfrac{1}{2} \le 1$: diverges.

(c) $p = 1.01 \gt 1$: converges. (Even a tiny bit more than $1$ is enough.)

(d) $n^2\sqrt{n} = n^2 \cdot n^{1/2} = n^{5/2}$, so $p = \tfrac{5}{2} \gt 1$: converges.

Example 2: Simplifying first

Does each series converge or diverge?

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{\sqrt{n}}{n^2}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \frac{3}{n\sqrt[3]{n}}$
• (c) $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{n^3}$

Solution.

(a) $\dfrac{n^{1/2}}{n^2} = \dfrac{1}{n^{3/2}}$, a p-series with $p = \tfrac{3}{2} \gt 1$: converges.

(b) $\dfrac{3}{n \cdot n^{1/3}} = 3 \cdot \dfrac{1}{n^{4/3}}$, a constant times a p-series with $p = \tfrac{4}{3} \gt 1$: converges.

(c) $\dfrac{n^2}{n^3} = \dfrac{1}{n}$, the harmonic series: diverges.

Example 3: Why the harmonic series diverges, without calculus

Show that the 16th partial sum of the harmonic series is at least $3$, and explain why the partial sums grow without bound.

Solution. Group the terms in blocks that double in length, and replace each term by the smallest term in its block:

$$\begin{aligned} S_{16} &= 1 + \frac{1}{2} + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \dots + \frac{1}{8}\right) + \left(\frac{1}{9} + \dots + \frac{1}{16}\right) \\ &\ge 1 + \frac{1}{2} + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} \cdot 4\right) + \left(\frac{1}{16} \cdot 8\right) \\ &= 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} = 3 \end{aligned}$$

(The actual value is $S_{16} \approx 3.381$.) Each new block adds at least another $\tfrac{1}{2}$, so $S_{2^k} \ge 1 + \tfrac{k}{2}$. That grows without bound, so the harmonic series diverges.

Example 4: A shifted p-series

Does $\displaystyle\sum_{n=1}^{\infty} \frac{4}{(n + 2)^3}$ converge or diverge?

Solution. Write out the terms: $\dfrac{4}{3^3} + \dfrac{4}{4^3} + \dfrac{4}{5^3} + \dots$ This is $4\displaystyle\sum_{m=3}^{\infty} \frac{1}{m^3}$, a constant times a p-series with $p = 3$ and its first two terms removed. Removing finitely many terms doesn’t affect convergence, so it converges.

Common mistakes

Thinking p = 1 converges. The rule is $p \gt 1$, strictly. The harmonic series ($p = 1$) diverges, and AP questions love to test exactly this case.

Mixing up p-series and geometric series. $\sum \frac{1}{n^2}$ is a p-series ($p = 2$, converges). $\sum \frac{1}{2^n}$ is geometric ($r = \tfrac{1}{2}$, converges). $\sum \frac{1}{n^{1/2}}$ is a p-series that diverges, while $\sum \left(\tfrac{1}{2}\right)^n$ converges. Check whether $n$ is in the base or the exponent.

Simplifying the powers wrongly. $n\sqrt{n} = n^{3/2}$, not $n^{1/2}$, and $\frac{1}{\sqrt[3]{n^2}} = \frac{1}{n^{2/3}}$. Write every root as a fractional exponent before deciding.

Saying the sum is 1/(p − 1). That’s the value of the integral. The sum of $\sum \frac{1}{n^2}$ is $\frac{\pi^2}{6}$, not $1$. On the AP exam you’ll only be asked whether a p-series converges, not for its sum.

Believing “terms go to 0, so it converges.” The harmonic series is the standard counterexample. Its terms go to $0$ and it still diverges.

Practice

1. (Warm-up) Does each series converge or diverge?

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^5}$
• (b) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{0.5}}$
• (c) $\displaystyle\sum_{n=1}^{\infty} n^{-1}$

Solution

(a) $p = 5 \gt 1$: converges.

(b) $p = 0.5 \le 1$: diverges.

(c) $n^{-1} = \frac{1}{n}$, the harmonic series ($p = 1$): diverges.

2. (Warm-up) Does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}}$ converge or diverge?

Solution

$\dfrac{1}{\sqrt[3]{n}} = \dfrac{1}{n^{1/3}}$, a p-series with $p = \tfrac{1}{3} \le 1$, so it diverges.

3. (Warm-up) Write $\dfrac{n^2}{n^{4.5}}$ as a single power and decide whether $\displaystyle\sum_{n=1}^{\infty} \frac{n^2}{n^{4.5}}$ converges.

Solution

$\dfrac{n^2}{n^{4.5}} = \dfrac{1}{n^{2.5}}$, a p-series with $p = 2.5 \gt 1$, so it converges.

4. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \frac{n + \sqrt{n}}{n^3}$ converge or diverge?

Solution

Split the fraction:

$$\frac{n + \sqrt{n}}{n^3} = \frac{1}{n^2} + \frac{1}{n^{5/2}}$$

Both $\sum \frac{1}{n^2}$ ($p = 2$) and $\sum \frac{1}{n^{5/2}}$ ($p = \tfrac{5}{2}$) converge, and the sum of two convergent series converges. So the series converges.

5. (Core) Does $\displaystyle\sum_{n=1}^{\infty} \left(\frac{1}{n^2} - \frac{1}{n}\right)$ converge or diverge? Explain.

Solution

It diverges. Suppose $\sum \left(\frac{1}{n^2} - \frac{1}{n}\right)$ converged. Then

$$\sum \frac{1}{n} = \sum \frac{1}{n^2} - \sum \left(\frac{1}{n^2} - \frac{1}{n}\right)$$

would be a difference of two convergent series, so it would converge. But the harmonic series diverges. So the original series must diverge. In short: convergent plus divergent is always divergent.

6. (Core) For which values of $k$ does $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^{2k - 3}}$ converge?

Solution

It’s a p-series with $p = 2k - 3$. It converges when $p \gt 1$:

$$2k - 3 \gt 1 \quad\Rightarrow\quad k \gt 2$$

7. (Core) Given that $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$, find the exact value of each sum.

• (a) $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n)^2}$
• (b) $\displaystyle\sum_{n=3}^{\infty} \frac{1}{n^2}$

Solution

(a) $\dfrac{1}{(2n)^2} = \dfrac{1}{4} \cdot \dfrac{1}{n^2}$, so the sum is $\dfrac{1}{4} \cdot \dfrac{\pi^2}{6} = \dfrac{\pi^2}{24}$.

(b) Remove the first two terms, $1$ and $\tfrac{1}{4}$:

$$\frac{\pi^2}{6} - 1 - \frac{1}{4} = \frac{\pi^2}{6} - \frac{5}{4}$$

8. (Challenge) Use Question 7 to find $\displaystyle\sum_{n=1}^{\infty} \frac{1}{(2n - 1)^2} = 1 + \frac{1}{9} + \frac{1}{25} + \dots$

Solution

The terms of $\sum \frac{1}{n^2}$ split into odd $n$ and even $n$. The even-$n$ terms are $\sum \frac{1}{(2n)^2} = \frac{\pi^2}{24}$. So the odd-$n$ terms add up to

$$\frac{\pi^2}{6} - \frac{\pi^2}{24} = \frac{4\pi^2 - \pi^2}{24} = \frac{\pi^2}{8}$$

(Splitting like this is fine because all the terms are positive and the series converges.)

9. (Challenge) Using $S_{2^k} \ge 1 + \tfrac{k}{2}$ from Example 3, find a number of terms $n$ that guarantees the harmonic series’ partial sum $S_n$ is at least $6$.

Solution

We need $1 + \tfrac{k}{2} \ge 6$, so $k \ge 10$. Then $n = 2^{10} = 1024$ terms are enough: $S_{1024} \ge 6$.

That’s a guarantee, not the smallest possible $n$. (A computer shows $S_n$ first passes $6$ at $n = 227$.) Either way, the sum grows very slowly.