## Easy Proof of the Divergence of the Harmonic Series

The Harmonic Series is a really good counterexample to the intuition that “any series that tends to 0 converges”. Let’s see why.

How do you determine if a series, particularly infinite ones,  converges or diverges? An intuitive guess would be something along the lines of this:

An Infinite Series $$S = sum_{n=1}^{\infty}\alpha_n$$ converges if $$\lim_{n \rightarrow \infty} \alpha_n = 0$$

, an admittedly confusing way of saying that the further you go down the series, the closer you get to zero, and you can get the series as close to zero as possible, by going down far enough down the series. For example, the series $$\alpha_n = (\frac{1}{2})^n$$ converges, because as you increase $$n$$, $$\alpha_n$$ gets really close to zero.

For example:

The series

$$\alpha_n = (\frac{1}{2})^n$$ converges, and so does

$$\gamma_n = (\frac{1}{n!})$$ (in fact, to the constant $$e$$, and so does

$$\beta_n = (-1)^n\frac{1}{n}$$.

But there is one item missing from the list…

## Proof of the Divergence of the Harmonic Series

Well, that’s also called the Harmonic Series (wiki), given to the series of the reciprocals of the Natural numbers.

Spoilers: it DIVERGES!

Here’s an elegant little proof (by contradiction, as usual) that demonstrates this:

Suppose that the sum $$\sum_{n=1}^{\infty}\frac{1}{n}$$ converges, and converges to $$S$$. Then, that can be restated:

$$S = \sum_{n=1}^{\infty}\frac{1}{n}$$.

If we take every other fraction, and increase its denominator (decreasing its value)

$$\frac{1}{n} \geq \frac{1}{n+1}$$, for $$n \in \mathbb{N}$$.

Here is are the first few items of the series, explicitly:

$$S = \sum_{n=1}^{\infty}\frac{1}{n} = \frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4} +\frac{1}{5} + \frac{1}{6} +\frac{1}{7} +…$$,

making the replacements (to $$\frac{1}{3}, \frac{1}{5}, \frac{1}{7}… \frac{1}{2n+1}$$, for $$n \in \mathbb{N}$$):

Replacing the elements with something smaller obviously makes the sum smaller, thus we can write:

$$S = \sum_{n=1}^{\infty}\frac{1}{n} > \frac{1}{1} +\frac{1}{2} +(\frac{1}{4} +\frac{1}{4}) +(\frac{1}{6} + \frac{1}{6}) +(\frac{1}{8} + \frac{1}{8}) + …$$

,if you add the fractions together, that comes out to:

$$= \frac{1}{1} +\frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + …$$,

which come out to

$$= \frac{1}{2} + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + …)$$

Substituting S for the part after the half, we obtain

$$S = \sum_{n=1}^{\infty}\frac{1}{n} > \frac{1}{2} + (\frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + …) = \frac{1}{2} + S$$,

which, if we take the front and ending terms, becoming the absurd inequality:

$$S > S + \frac{1}{2}$$.

Which is a contradiction.

### Conclusion:

The result we just proved means one thing:

The Harmonic Series does not converge to a single value.

What does this mean? Well, we can make the series sum to be larger than any number, provided we sum enough terms! Although it may be counter-intuitive, the proof is above. And this can lead to some really counter-intuitive results:

Let’s see one from WikiPedia:

If I had an unlimited amount of blocks, and a table, I can stack the blocks an arbitrary distance from the table, provided I use enough blocks.

This is pretty easy: Just stack the first block with a half sticking out, then the second with a third sticking out, the third with a fourth sticking out, and so on. The amount sticking out will just be the series above, which tends to infinity, albeit very, very slowly.