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10.5 - Harmonic Series and p-Series

Definition of a p-Series

A p-series is an infinite series of the form:
∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}
where pp is a positive constant.
Written out:
∑n=1∞1np=11p+12p+13p+14p+⋯\sum_{n=1}^{\infty} \frac{1}{n^p} = \frac{1}{1^p} + \frac{1}{2^p} + \frac{1}{3^p} + \frac{1}{4^p} + \cdots
The defining characteristic: the variable nn is in the base, and the constant pp is the exponent.

Convergence Test for p-Series

Everything depends on the value of pp:
∑n=1∞1np{CONVERGESif p>1DIVERGESif 0<p≤1\boxed{\sum_{n=1}^{\infty} \frac{1}{n^p} \begin{cases} \text{CONVERGES} & \text{if } p > 1 \\ \text{DIVERGES} & \text{if } 0 < p \leq 1 \end{cases}}
10.5 - Harmonic Series and p-Series
Memory aid
When pp is large (greater than 1), terms like 1n500\frac{1}{n^{500}} shrink to nothing almost immediately. They approach 0 fast enough that the sum settles on a finite value.
When pp is small (1 or less), the terms don’t shrink fast enough, and the sum grows without bound.
Why the cutoff is exactly p=1p = 1: it comes from the integral test (10.4). For p>0p > 0, f(x)=1xpf(x) = \frac{1}{x^p} is positive, continuous, and decreasing for x≥1x \geq 1, and
∫1b1xp dx=b1−p−11−p(p≠1)\int_1^{b} \frac{1}{x^p}\,dx = \frac{b^{1-p} - 1}{1 - p} \quad (p \neq 1)
  • p>1p > 1: b1−p→0b^{1-p} \to 0, so the integral equals 1p−1\frac{1}{p-1}, a finite number. The series converges.
  • p<1p < 1: b1−p→∞b^{1-p} \to \infty, so the integral diverges, and so does the series.
  • p=1p = 1: ∫1b1x dx=ln⁡b→∞\int_1^{b} \frac{1}{x}\,dx = \ln b \to \infty, so the harmonic series diverges.

The Harmonic Series

The harmonic series is the special case where p=1p = 1:
∑n=1∞1n=1+12+13+14+⋯\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \cdots
The harmonic series diverges. This is a famous result, because even though the terms approach zero, they don’t shrink fast enough to give a finite sum.
The alternating harmonic series (coming in a later lesson) behaves differently and actually converges.
∑n=1∞(−1)n+1n=1−12+13−14+15−⋯\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots
Same sizes as the harmonic series, but the signs alternate, so each term partly cancels the one before it. It converges by the Alternating Series Test (10.7), and its sum is ln⁡2≈0.693\ln 2 \approx 0.693 (see 10.10).

Examples

Example 1: a clear p-series
Does ∑n=1∞1n4\sum_{n=1}^{\infty} \frac{1}{n^4} converge or diverge?
This is a p-series with p=4p = 4. Since 4>14 > 1, it converges.
Example 2: p-series with a radical
Does ∑n=1∞1n3\sum_{n=1}^{\infty} \frac{1}{\sqrt[3]{n}} converge or diverge?
Rewrite with exponents:
1n3=1n1/3=n−1/3\frac{1}{\sqrt[3]{n}} = \frac{1}{n^{1/3}} = n^{-1/3}
This is a p-series with p=13p = \frac{1}{3}. Since 13<1\frac{1}{3} < 1, it diverges.
Example 3: p-series vs. geometric series
Does ∑n=1∞14n\sum_{n=1}^{\infty} \frac{1}{4^n} converge or diverge?
Careful, this is not a p-series. Rewrite:
14n=(14)n\frac{1}{4^n} = \left(\frac{1}{4}\right)^n
That’s a geometric series with r=14r = \frac{1}{4}. Since ∣r∣<1|r| < 1, it converges by the geometric series test.
Type
Form
Where nn lives
Test
p-series
1np\frac{1}{n^p}
Base
p>1p > 1 → converges
Geometric
(1a)n\left(\frac{1}{a}\right)^n
Exponent
∣r∣<1|r| < 1 → converges

Critical Comparison: p-Series vs. Geometric

p-series: ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}, where nn is the base. Converges when p>1p > 1.
Geometric: ∑n=0∞arn\sum_{n=0}^{\infty} ar^n, where nn is the exponent. Converges when ∣r∣<1|r| < 1.
The rules point in opposite directions, so don’t mix them up:
  • p-series: the exponent value needs to be greater than 1 to converge
  • Geometric: the base value (rr) needs to be less than 1 in absolute value to converge

Finding Values That Make a Series Converge

Example 4: linear expression in the exponent
For what values of kk does ∑n=1∞1n3k−2\sum_{n=1}^{\infty} \frac{1}{n^{3k-2}} converge?
This is a p-series with p=3k−2p = 3k - 2. For convergence we need p>1p > 1:
3k−2>1⇒3k>3⇒k>13k - 2 > 1 \quad \Rightarrow \quad 3k > 3 \quad \Rightarrow \quad k > 1
The series converges when k>1k > 1 and diverges when k≤1k \leq 1.
Example 5: product in the denominator
For what values of kk does ∑n=1∞1n⋅n5k\sum_{n=1}^{\infty} \frac{1}{n \cdot n^{5k}} converge?
Step 1: simplify with exponent rules
1n⋅n5k=1n1+5k\frac{1}{n \cdot n^{5k}} = \frac{1}{n^{1 + 5k}}
This is a p-series with p=1+5kp = 1 + 5k.
Step 2: set up the inequality
1+5k>1⇒5k>0⇒k>01 + 5k > 1 \quad \Rightarrow \quad 5k > 0 \quad \Rightarrow \quad k > 0
The series converges when k>0k > 0 and diverges when k≤0k \leq 0.
Example 6: denominator with a sum
For what values of kk does ∑n=1∞nn3k+7\sum_{n=1}^{\infty} \frac{n}{n^{3k} + 7} converge?
Step 1: factor out the dominant term
nn3k+7=nn3k(1+7n3k)=1n3k−1(1+7n3k)\frac{n}{n^{3k} + 7} = \frac{n}{n^{3k}\left(1 + \frac{7}{n^{3k}}\right)} = \frac{1}{n^{3k-1}\left(1 + \frac{7}{n^{3k}}\right)}
Step 2: analyze as n→∞n \to \infty. The term 7n3k→0\frac{7}{n^{3k}} \to 0, so:
1n3k−1(1+7n3k)≈1n3k−1\frac{1}{n^{3k-1}\left(1 + \frac{7}{n^{3k}}\right)} \approx \frac{1}{n^{3k-1}}
Step 3: apply the p-series test. This behaves like a p-series with p=3k−1p = 3k - 1:
3k−1>1⇒3k>2⇒k>233k - 1 > 1 \quad \Rightarrow \quad 3k > 2 \quad \Rightarrow \quad k > \frac{2}{3}
The series converges when k>23k > \frac{2}{3} and diverges when k≤23k \leq \frac{2}{3}.

Summary of Convergence Tests So Far

1. Geometric series test
∑n=0∞arn\sum_{n=0}^{\infty} ar^n
  • Converges if ∣r∣<1|r| < 1, to the sum a1−r\frac{a}{1-r}
  • Diverges if ∣r∣≥1|r| \geq 1
2. nth term test for divergence
If lim⁡n→∞an≠0  ⟹  the series DIVERGES\text{If } \lim_{n \to \infty} a_n \neq 0 \implies \text{the series DIVERGES}
  • Limit is 0 → no conclusion (it may converge or diverge)
  • Limit isn’t 0 → definitely diverges
3. Integral test
If f(x)f(x) is positive, continuous, and decreasing for x≥kx \geq k, then ∑n=k∞an\sum_{n=k}^{\infty} a_n and ∫k∞f(x) dx\int_{k}^{\infty} f(x) \, dx both converge or both diverge together.
4. p-series test (new)
∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}
  • Converges if p>1p > 1
  • Diverges if 0<p≤10 < p \leq 1

Strategy for Choosing a Test

Step 1: Try the nth term test first. It’s the fastest way to show divergence.
Step 2: Check if it’s geometric (look for rnr^n).
Step 3: Check if it’s a p-series (look for 1np\frac{1}{n^p}).
Step 4: Consider the integral test if the function is easy to integrate.
Step 5: Move on to other tests (coming in later lessons).

Common p-Series

Series
pp value
Result
∑1n\sum \frac{1}{n}
p=1p = 1
Diverges (harmonic)
∑1n1/2\sum \frac{1}{n^{1/2}}
p=0.5p = 0.5
Diverges
∑1n1.01\sum \frac{1}{n^{1.01}}
p=1.01p = 1.01
Converges
∑1n2\sum \frac{1}{n^2}
p=2p = 2
Converges
∑1n3\sum \frac{1}{n^3}
p=3p = 3
Converges
∑1n\sum \frac{1}{\sqrt{n}}
p=0.5p = 0.5
Diverges

Key Takeaways

  • p-series form: ∑n=1∞1np\sum_{n=1}^{\infty} \frac{1}{n^p}, where pp is a positive constant
  • Convergence rule: converges if p>1p > 1, diverges if 0<p≤10 < p \leq 1
  • Harmonic series: the special case p=1p = 1, and it always diverges
  • Critical value: p=1p = 1 is the boundary between converging and diverging
  • Variable position matters: p-series has nn in the base, geometric has nn in the exponent
  • Opposite rules: p-series needs p>1p > 1, geometric needs ∣r∣<1|r| < 1
  • Simplify first: use exponent rules to rewrite messy expressions as 1np\frac{1}{n^p}
  • Denominators with sums: factor out the dominant term and analyze the limit
  • Test hierarchy: nth term → geometric → p-series → integral → others
  • No integration needed: the p-series test is instant once you spot pp

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