Harmonic Series for MATH 141
Exam Relevance for MATH 141
The harmonic series is a MATH 141 reference point for comparisons. Know it diverges.
The Most Famous Divergent Series
The harmonic series is:
$$\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots$$
It diverges — even though the terms go to zero!
This is the classic example showing that $a_n \to 0$ is NOT enough to guarantee convergence.
Why "Harmonic"?
The name comes from music. In acoustics, the harmonic frequencies of a vibrating string are multiples of a fundamental frequency: $f, 2f, 3f, 4f, \ldots$
The wavelengths are proportional to $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$ — the harmonic series!
The Book Stacking Problem
Here's something you can try at home: How far can you make a stack of books overhang the edge of a table?
With 1 book, you can overhang $\frac{1}{2}$ a book length (center of mass at the edge).
With 2 books, the top book overhangs $\frac{1}{2}$, and you shift the whole stack by $\frac{1}{4}$. Total: $\frac{1}{2} + \frac{1}{4} = \frac{3}{4}$.
With $n$ books, the maximum overhang is:
$$\text{Overhang} = \frac{1}{2}\left(1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}\right) = \frac{H_n}{2}$$
where $H_n$ is the $n$th harmonic number.
The punchline: Since the harmonic series diverges, you can achieve infinite overhang with enough books! You could theoretically build a stack that extends a mile past the table edge.
| Books | Overhang (book lengths) |
|---|---|
| 4 | 1.04 (just past 1 book!) |
| 31 | 2.01 |
| 227 | 3.00 |
| 1,674 | 4.00 |
It takes 31 books just to overhang 2 book lengths. The divergence is real, but painfully slow!
Why this matters: This isn't just a party trick. It shows that "diverges slowly" still means "diverges" — small contributions, repeated forever, eventually add up to anything you want.
Why Does It Diverge?
The terms shrink, so why doesn't the sum stay finite? Here's the classic proof by grouping:
$$1 + \frac{1}{2} + \underbrace{\frac{1}{3} + \frac{1}{4}}_{> \frac{1}{4} + \frac{1}{4} = \frac{1}{2}} + \underbrace{\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}}_{> 4 \cdot \frac{1}{8} = \frac{1}{2}} + \underbrace{\frac{1}{9} + \cdots + \frac{1}{16}}_{> 8 \cdot \frac{1}{16} = \frac{1}{2}} + \cdots$$
Each group of terms sums to more than $\frac{1}{2}$.
Since we're adding $\frac{1}{2}$ infinitely many times:
$$\sum \frac{1}{n} > 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \cdots = \infty$$
The Harmonic Series is P-Series with p = 1
Remember the p-series rule:
$$\sum \frac{1}{n^p} \text{ converges if } p > 1, \text{ diverges if } p \leq 1$$
The harmonic series is $\sum \frac{1}{n^1}$, so $p = 1$.
Since $p = 1$ is NOT greater than 1, it diverges.
The harmonic series sits exactly at the boundary between convergence and divergence.
How Slowly Does It Diverge?
The harmonic series diverges incredibly slowly. The partial sums grow like $\ln n$:
$$S_n = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n} \approx \ln n + \gamma$$
where $\gamma \approx 0.5772$ is the Euler-Mascheroni constant.
| $n$ | $S_n$ (approx) |
|---|---|
| 10 | 2.93 |
| 100 | 5.19 |
| 1,000 | 7.49 |
| 1,000,000 | 14.39 |
| $10^{100}$ | ≈ 230 |
To get $S_n > 100$, you need roughly $n > e^{100} \approx 10^{43}$ terms!
Variations of the Harmonic Series
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n+5}$ converges or diverges.
This is just the harmonic series starting later:
$$\frac{1}{6} + \frac{1}{7} + \frac{1}{8} + \cdots$$
Removing a finite number of terms doesn't change divergence.
$$\boxed{\text{Diverges}}$$
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{3}{n}$ converges or diverges.
$$\sum_{n=1}^{\infty} \frac{3}{n} = 3 \sum_{n=1}^{\infty} \frac{1}{n} = 3 \cdot (\text{divergent}) = \text{divergent}$$
A constant multiple of a divergent series still diverges.
$$\boxed{\text{Diverges}}$$
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{1}{2n}$ converges or diverges.
$$\sum_{n=1}^{\infty} \frac{1}{2n} = \frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n} = \frac{1}{2} \cdot (\text{divergent}) = \text{divergent}$$
$$\boxed{\text{Diverges}}$$
This is $\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + \cdots$ — half the harmonic series, still divergent!
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{1}{2n-1}$ converges or diverges.
This is $1 + \frac{1}{3} + \frac{1}{5} + \frac{1}{7} + \cdots$
Notice: (odd terms) + (even terms) = harmonic series
If both converged, harmonic would converge. But harmonic diverges!
Since the even-denominator series $\sum \frac{1}{2n}$ diverges (Example 3), and:
$$\sum \frac{1}{n} = \sum \frac{1}{2n-1} + \sum \frac{1}{2n}$$
The odd-denominator series must also diverge.
$$\boxed{\text{Diverges}}$$
Alternating Harmonic Series (Preview)
Something magical happens when we alternate signs:
$$\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n} = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \cdots = \ln 2$$
This converges! Alternating signs can turn a divergent series into a convergent one.
(We'll study this more in the Alternating Series lesson.)
Using Harmonic Series for Comparison
The harmonic series is a benchmark for showing divergence.
Problem: Determine if $\displaystyle\sum_{n=1}^{\infty} \frac{1}{n+1}$ converges or diverges.
Compare to the harmonic series. For all $n \geq 1$:
$$\frac{1}{n+1} \text{ behaves like } \frac{1}{n}$$
More precisely: $\displaystyle\lim_{n \to \infty} \frac{\frac{1}{n+1}}{\frac{1}{n}} = \lim_{n \to \infty} \frac{n}{n+1} = 1$
By the Limit Comparison Test with the harmonic series (divergent):
$$\boxed{\text{Diverges}}$$
Problem: Determine if $\displaystyle\sum_{n=2}^{\infty} \frac{1}{\ln n}$ converges or diverges.
For $n \geq 2$: $\ln n < n$, so $\frac{1}{\ln n} > \frac{1}{n}$
Since $\sum \frac{1}{n}$ diverges and our terms are LARGER:
$$\boxed{\text{Diverges by Direct Comparison}}$$
Summary: Harmonic Series Facts
| Statement | True/False |
|---|---|
| $\sum \frac{1}{n}$ converges | ❌ False |
| $\lim_{n \to \infty} \frac{1}{n} = 0$ | ✓ True |
| $a_n \to 0$ implies convergence | ❌ False |
| Harmonic series is a p-series with $p = 1$ | ✓ True |
| Alternating harmonic series converges | ✓ True |
| Partial sums grow like $\ln n$ | ✓ True |
Common Mistakes and Misunderstandings
❌ Mistake: Thinking terms going to zero means convergence
Wrong: "The terms $\frac{1}{n} \to 0$, so $\sum \frac{1}{n}$ must converge."
Why it's wrong: This is the most common misconception in series! The harmonic series is the perfect counterexample. Terms going to zero is necessary but not sufficient for convergence.
Correct: $a_n \to 0$ only means the series might converge. You still need to test it.
❌ Mistake: Confusing harmonic with geometric
Wrong: "$\sum \frac{1}{n}$ is geometric with $r = \frac{1}{n}$."
Why it's wrong: In a geometric series, $r$ is a constant. Here, the denominator $n$ changes with each term.
Correct: Geometric: $\sum ar^n$ (constant ratio). Harmonic: $\sum \frac{1}{n}$ (p-series with $p = 1$).
❌ Mistake: Thinking "half the harmonic series" converges
Wrong: "$\sum \frac{1}{2n}$ is only half the terms, so maybe it converges."
Why it's wrong: $\sum \frac{1}{2n} = \frac{1}{2} \sum \frac{1}{n}$, which is half of infinity — still infinity!
Correct: Constant multiples of divergent series still diverge.
❌ Mistake: Not recognizing harmonic-like series
Wrong: "$\sum \frac{1}{n+100}$ looks different from harmonic, so I need a special test."
Why it's wrong: This is just the harmonic series missing its first 100 terms. Removing finitely many terms doesn't change convergence/divergence.
Correct: $\sum \frac{1}{n+k}$ diverges for any constant $k$ — it's essentially harmonic.
Harmonic Series
The harmonic series diverges even though its terms approach 0. It's a p-series with p = 1, exactly at the boundary.
Variables:
- $1/n$:
- the nth term, which approaches 0
- $S_n$:
- partial sums grow like ln(n)
Harmonic Partial Sum Approximation
The partial sums of the harmonic series grow logarithmically. γ ≈ 0.5772 is the Euler-Mascheroni constant.
Variables:
- $S_n$:
- sum of first n terms
- $γ$:
- Euler-Mascheroni constant ≈ 0.5772
Alternating Harmonic Series
Unlike the regular harmonic series, the alternating harmonic series converges (to ln 2). Alternating signs can turn divergence into convergence!
Variables:
- $(-1)^{n+1}$:
- alternates: +1, -1, +1, -1, ...
- $ln 2$:
- ≈ 0.693
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