Why does the $ p $-series diverge when $ p = 1 $?

Question

We’re currently working with $ p $-series in my Calculus class, and I’ve fallen for the apparently common misconception that the infinite sum $ \displaystyle \sum_{n=1}^{\infty} \frac{1}{n^{p}} $, where $ p = 1 $, should converge, whereas in reality, it diverges. Can anyone demonstrate or link me to a proof or explanation of why this is so?

Answer 1

You are considering the series $$1+\frac12+\frac13+\frac14+\cdots$$ One direct way to see that this diverges without any of the standard calculus tests goes as follows. Start grouping terms together like this: $$1+\overbrace{\frac12}^{2^0\text{ terms}}+\overbrace{\frac13+\frac14}^{2^1\text{ terms}}+\overbrace{\frac15+\frac16+\frac17+\frac18}^{2^2\text{ terms}}+\cdots$$ This is greater than $$1+\overbrace{\frac12}^{2^0\text{ terms}}+\overbrace{\frac14+\frac14}^{2^1\text{ terms}}+\overbrace{\frac18+\frac18+\frac18+\frac18}^{2^2\text{ terms}}+\cdots$$ which is equal to $$1+\overbrace{\frac12}+\overbrace{\frac12}+\overbrace{\frac12}+\cdots$$ which is clearly a divergent sum. To formalize this, you would show that partial sums of the harmonic series have $s_{2^n}\ge1+\frac{n}{2}$.

Answer 2

$$N=1+\underbrace{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}}_{>\,1}+\underbrace{\frac{1}{5}+\frac{1}{6}+\frac{1}{7}}_{>\,1/2}+\underbrace{\frac{1}{8}+\frac{1}{9}+\frac{1}{10}}_{>\,1/3}\cdots>1+N$$

$N$ cannot be finite!

Answer 3

To answer your question, the integral test says that if $a_n$ is a positive decreasing sequence, and there exists a decreasing positive $f$ such that $f(n)=a_n$ for each $n$, then $$\sum_{n=1}^\infty a_n$$ converges if and only if $$\int_1^\infty f(x) dx$$ does. You have correctly found that with $f(x)=\frac 1 x $ and $a_n=\frac 1 n$ we can prove divergence. Using the integral test, you can also show that any $p$ series converges for $p>1$ and diverges for $p\leq 1$.

Another elegant proof is the following. The $1$-series (a.k.a. Harmonic series), goes as follows. Let

$$H_{n}=\sum_{k=1}^n \frac 1 k $$

Then $$\eqalign{ & {H_2} = \sum\limits_{k = 1}^2 {{1 \over k}} = 1 + 1 \cdot {1 \over 2} \cr & {H_{{2^2}}} = \sum\limits_{k = 1}^4 {{1 \over k}} = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} > 1 + {1 \over 2} + {2 \over 4} = 1 + 2 \cdot {1 \over 2} \cr & {H_{{2^3}}} = \sum\limits_{k = 1}^8 {{1 \over k}} = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7} + {1 \over 8} > 1 + {1 \over 2} + {2 \over 4} + {4 \over 8} = 1 + 3 \cdot {1 \over 2} \cr & {H_{{2^4}}} = \sum\limits_{k = 1}^{16} {{1 \over k}} = 1 + {1 \over 2} + {1 \over 3} + {1 \over 4} + {1 \over 5} + {1 \over 6} + {1 \over 7} + {1 \over 8} + \cdots + {1 \over {16}} > 1 + {1 \over 2} + {2 \over 4} + {4 \over 8} + {8 \over {16}} = 1 + 4 \cdot {1 \over 2} \cr} $$

In general, we then show $${H_{{2^n}}} > 1 + {n \over 2}$$ which proves the Harmonic or $1$-series must diverge.

Answer 4

The various articles on Wikipedia and the like were helpful, so thank you for that, but I only truly felt confident in the answer once I did an integral test of it, which reveals that it tends to infinity.

$$\lim_{b\to \infty} \int_1^b \frac{dx}x=$$

$$\lim_{b\to \infty} \log b-\log 1=$$

$$\lim_{b\to \infty} \log b\to \infty$$

Pretty sure that's right - anyone know how to format that more elegantly?