The Harmonic Series

Consider the infinite sum:

1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \dots

This is called the harmonic series. Each term gets smaller and smaller, approaching zero.

Does this sum converge to a finite number, or diverge to infinity? If it converges, what is the sum? If it diverges, prove it.

Hint

Group the terms into blocks of increasing size: the first term alone, then the next term, then the next 2 terms, then the next 4 terms, then the next 8 terms, and so on.

Solution

Answer: The harmonic series diverges to infinity.

Let’s group the terms after the first two in blocks of increasing size:

1 + \frac{1}{2} + (\frac{1}{3} + \frac{1}{4}) + (\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}) + (\frac{1}{9} + \dots + \frac{1}{16}) + \dots

Each group has twice as many terms as the previous one. The first group has 2 terms, the next has 4, then 8, then 16, and so on.

Now let’s find a lower bound for each group. In the group $(\frac{1}{3} + \frac{1}{4})$ , both terms are at least $\frac{1}{4}$ , so the sum is at least $2 \times \frac{1}{4} = \frac{1}{2}$ .

In the group $(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8})$ , all four terms are at least $\frac{1}{8}$ , so the sum is at least $4 \times \frac{1}{8} = \frac{1}{2}$ .

In the group $(\frac{1}{9} + \dots + \frac{1}{16})$ , all eight terms are at least $\frac{1}{16}$ , so the sum is at least $8 \times \frac{1}{16} = \frac{1}{2}$ .

By the same reasoning, every group after the first two terms contributes at least $\frac{1}{2}$ to the total sum.

Since we have infinitely many such groups, each contributing at least $\frac{1}{2}$ , the total sum is at least:

1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \frac{1}{2} + \dots

This sum diverges to infinity. Thus, the harmonic series diverges.

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