Why do we say the harmonic series is divergent?

Question

If we have $\Sigma\frac{1}{n}$, why do we say it is divergent? Yes, it is constantly increasing, but after a certain point, $n$ will be so large that we will be certain of millions of digits. If we continue to let $n$ increase, we will end up with a number so large in the denominator that there will be an infinite amount of digits locked into place. What we would have left would just be an irrational number, correct? $\pi$ is an irrational number, but we still consider the value of that to be know. The common estimation of $\pi$ is 3.141592, and we can calculate it past 1,000,000 decimal places, so why can we just assume that we know the first few million places of the harmonic series, slap an irrational label on it, and call it a day? After all, the series $\Sigma\frac{1}{n^n}$ is convergent, and it basically does the same thing, it just gets there a lot faster.

I feel like argument has probably been made before, so I feel like there's probably a proof somewhere proving me wrong, if someone could point me to that.

Answer 1

The sum $\frac{1}{3}+\frac{1}{4}$ is $\gt \frac{1}{2}$.

The sum $\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}$ is $\gt \frac{1}{2}$. This is easy, we have $4$ terms, each $\ge \frac{1}{8}$, and all but one $\gt \frac{1}{8}$.

The sum $\frac{1}{9}+\frac{1}{10}+\cdots+\frac{1}{16}$ is $\gt \frac{1}{2}$. We have $8$ terms, each $\ge \frac{1}{16}$, and all but one $\gt \frac{1}{16}$.

The sum $\frac{1}{17}+\frac{1}{18}+\cdots +\frac{1}{32}$ is $\gt \frac{1}{2}$.

The sum of the terms $\frac{1}{33}$ to $\frac{1}{64}$ is $\gt \frac{1}{2}$.

The sum of the terms $\frac{1}{65}$ to $\frac{1}{128}$ is $\gt \frac{1}{2}$.

And so on.

Counting in the first two terms, if we add until the term $\frac{1}{4}$ our sum is $\gt 2$. If we add up to $\frac{1}{8}$, our sum is $\gt 2.5$. Adding to $\frac{1}{16}$ puts us beyond $3$. At $32$ terms, we are beyond $3.5$. At $64$ terms, we are beyond $4$. At $256$, we are beyond $5$. At $1024$, we are beyond $6$. At $4096$, we are beyond $7$.

Painfully slow! But if we are patient enough, after a while (a long long while) we will be beyond $10$, beyond $100$, beyond $1000$. But the universe may end first.

Remarks: On a calculator, the sum is finite! Because of roundoff, after a while we are just adding $0$.

The answer dealt with the series $\sum \frac{1}{n}$. It turns out that for any positive $\epsilon$, the series $\sum \frac{1}{n^{1+\epsilon}}$ converges. We can take for example $\epsilon=0.0001$. So one can say that $\sum \frac{1}{n}$ diverges extremely reluctantly, and that close neighbours converge.

Answer 2

No digits will be locked. It's diverging because given any number X, if you add together enough terms of the sum you will eventually exceed X. (Using finite number of terms.)

Answer 3

The number in the denominator would be large, but the numerator would be larger. "Infinitely larger", which is the whole point.

$$...123129.120931293...$$

(random example) is not an irrational number because of the infinitely many digits to the left. You can write the digits down (sort of), but they don't have a number associated with them this way. A real number is in fact often defined as the limit of a converging sequence, so a sequence that gets larger and larger can't be a real number by definition. Trying to assign a real number to $\sum_1^\infty\frac 1 n$ is like trying to give me the largest integer. Whatever answer you give, I can go higher, so the answer doesn't exist.

Answer 4

The idea with the harmonic series is that you can let $$ \sum_{n=1}^N \frac 1n $$ to be as large as you want. It's not because we "cannot compute it precisely enough" that we label it infinity ; it because it doesn't make sense to attribute a real number to its value in the sense of $$ \lim_{N \to \infty} \sum_{n=1}^N \frac 1n. $$ If you have ever seen the proof of the divergence of this sum, here it is : $$ \sum_{n=1}^{2^N} \frac 1n = 1 + \sum_{i=1}^{N} \sum_{n=2^{i-1}+1}^{2^i} \frac 1n \ge \sum_{i=1}^{N} \sum_{n=2^{i-1}+1}^{2^i}\frac 1{2^i} = \sum_{i=1}^{N} \frac 12 = \frac N2 $$ and letting $N \to \infty$ means that in particular the sum $$ \sum_{n=1}^K \frac 1n \ge \sum_{n=1}^{2^N} \frac 1n \ge \frac N2 $$ if we take $K$ such that $K > 2^N$. If you don't understand the formalism, the idea is that I just regrouped the terms like this : $$ 1 + \frac 12 + \frac 13 + \frac 14 + \frac 15 + \frac 16 + \frac 17 + \frac 18 + \dots \ge 1 + \frac 12 + (\frac 14 + \frac 14) + (\frac 18 + \frac 18 + \frac 18 + \frac 18) + \dots $$ and that by using $2^i$ terms each time, I can bound by below each term of, say $\frac 1{17},\frac 1{18}, \dots, \frac 1{32}$ by $\frac 1{32}$, and since there are $16$ such terms, the sum of them is $\frac 12$. And it keeps going on like this.

Hope that helps,

Answer 5

What you are trying to argue is that if $a_n$ is a sequence of non-negative numbers, and $a_n\to 0$ as $n\to\infty$ then the sum $\sum_{i=1}^\infty a_n$ ought to converge.

Unfortunately, that is not true. It would simplify a lot of math if it was true.

Have you seen the proof that it diverges?

Let $$s_m=\sum_{n=1}^m \frac 1 n$$

be the $m$th partial sum.

Then $$s_{2m}-s_{m} = \sum_{n=m+1}^{2m} \frac{1}{n} \geq \sum_{n=m+1}^{2m}\frac{1}{2m} = \frac{1}{2}.$$

In particular, you can show by induction that $s_{2^k} \geq \frac{k}{2}$. So $s_m$ can be made as large as you like.

Answer 6

The mistake you make here is the standard mistake many students make when they think convergence of $\sum a_n$ is the same as $\lim a_n =0$.

The so called Divergence test sais that if the first happens then the second must hold, but the converse is not true. And here is the reason why:

If $\lim a_n =0$, then $a_n$ becomes smaller and smaller, but when you calculate $\sum a_n$ you add MANY terms, and the sum of many small numbers can be anything. It is very similar to the type of limit $\infty \cdot 0$, you have infinitely many terms, all of them close to 0.... And anything can happen in this case.

For this series, what happens is that more and more terms will agree on the first "few" digits, but then KABOOM you get a term with new digits..The actual series, or more correctly, the sequence of partial sums $s_n$, goes to infinity very very slowly, but it grows to infinity...

$s_n$ actually behaves exactly like $\ln(n)$, and looking to this sequence migth clarify a little what happens: $\ln(n) \to \infty$ but $\frac{1}{n} \sim \ln(n+1)-\ln(n) \to 0$. Exactly like $\ln(n)$, $s_n$ grows to $\infty$, even if at each step it grows by $s_n-s_{n-1}$, an amount which goes to 0....

Answer 7

Consider the sum $$H_n = \sum_{k=1}^n\frac{1}{k}$$ for $n=1,2,\ldots$. These are called the harmonic numbers. You are interested in $\lim_{n\to\infty}H_n$.

Notice that $H_3 = 1 +\frac{1}{2}+\frac{1}{3}$ is just the area of the shaded region between $x=1$ and $x=4$ in the figure below. For any $n$ the area of the shaded region will be larger than the corresponding area under the red line, the graph of $y = \frac{1}{x}$. This implies $$\begin{eqnarray*} H_n &=& \sum_{k=1}^n \frac{1}{k} \\ &>& \int_1^{n+1} \frac{dx}{x} \\ &>& \int_1^n \frac{dx}{x}. \end{eqnarray*}$$ Similarly, an upper bound can be found, $$\begin{eqnarray*} H_n &<& 1+\int_{2}^{n+1} \frac{dx}{x-1} \\ &=& 1+ \int_1^n \frac{dx}{x}. \end{eqnarray*}$$ Therefore, $$\log n < H_n < \log n + 1.$$ The lower bound implies $\lim_{n\to\infty}H_n = \infty$, that is, the sum grows without bound. In addition we find $H_n \sim \log n \, (n\to\infty)$, that is, $H_n$ behaves like $\log n$ for large $n$.

Your intuition that $H_n$ grows very slowly is spot on. Roughly, if we want the sum to be some number $N$ we need about $e^N$ terms. For example, $H_n \simeq 185{}^\dagger$ for $n \simeq e^{185}$. That is, we need as many terms as there are atoms in the universe for the sum to reach $185$.

Figure 1. Illustration of the integral test in calculus by Jim Belk.

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_{${}^\dagger$ Actually, $H_n \simeq 185.577$.}