Divergence test (2024)

*The divergence test can determine whether a series diverges, and if it does diverge, it can't possibly converge. But the divergence test is not a test for convergence.

A series can pass the divergence test (appear to converge), and still diverge when subjected to another test.

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Converse

In logic, we can encounter a case in which, if a thing (we'll call it P) is true, then something else (called Q) must be true. We write that as P → Q.

The converse of that statement is Q → P. The converse of a proposition like P → Q might be true, but it isn't necessarily true.

That is, just because A proves B, doesn't mean that B proves A.

We start with the n^th partial sum of a series, S_n,

$$S_n = (a_1 + a_2 + \dots + a_{n - 1}) + a_n$$

Note that we've kept the a_n term on the outside to give a sum of a_n and a sub-series of S_n, S_n-1:

Now we'll substitute S_n-1 for that sum in parentheses, then rearrange to solve for a_n:

$$ \begin{align} S_n &= S_{n - 1} + a_n \\ \\ a_n = S_n - S_{n - 1} \end{align}$$

Now we have an expression for a_n of which we can take advantage. Here is the substance of the proof:

$$ \begin{align} If \: \sum_{n = 1}^{\infty} a_n &= S \: converges, \: then \\ \\ \lim_{n\to\infty} a_n &= \lim_{n\to\infty} (S_n - S_{n - 1}) \\ \\ &= \lim_{n\to\infty} S_n - \lim_{n\to\infty} S_{n - 1} \\ \\ &= S - S = 0 \end{align}$$

In that last step we note that n-1 → ∞ as n → ∞, so the limits are the same (S). Another way to think of it is that for large enough partial sums of an infinite series, the difference between a_n and a_n-1 is small when n is large. Now if a_n does not converge to zero, then the sum of the terms, a_n, cannot converge.

Show that $\sum_{n = 1}^{\infty} \frac{n}{2n + 1}$ diverges.

Solution: The divergence test simply asks whether the n^th term of the series has a non-zero limit. If it does, the series diverges. The limit can be found using L'Hopital's rule:

$$\lim_{n\to\infty} \frac{n}{2n + 1} = \lim_{n\to\infty} \frac{1}{2} = \frac{1}{2} \ne 0$$

The limit is not zero. That means

$$\sum_{n = 1}^{\infty} \frac{n}{2n + 1} = \frac{1}{3} + \frac{2}{5} + \frac{3}{7} + \dots$$

This series does not reach some finite limit, but continues to grow. For very large n, the series looks like ½ + ½ + ½ ...

This series will not be convergent, and is therefore not very useful as a series to represent some finite-valued function.

This plot shows the size of each term for n = 1 to 11 and how the series grows as terms are added. A convergent series would have an asymptote that the sum would approach as n → ∞.

If an infinite series fails the divergence test, we're done with it. It is not a convergent series, and therefore cannot be used to represent a finite-valued function.

Example 2

The terms tend to zero, but the series diverges

Show that the series $\sum_{n = 1}^{\infty}$ diverges.

Solution: This series will show that just because the divergence test shows that the series does not diverge, doesn't mean we can't show that it diverges another way. Here are the first few terms of the series:

$$\sum_{n = 1}^{\infty} \frac{1}{\sqrt{n}} = \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \dots$$

Clearly the limit of the terms as n → ∞ is zero:

$$\lim_{n\to\infty} \frac{1}{\sqrt{n}} = 0$$

So it looks like this series converges: The limit of the terms is zero, so it doesn't fail the divergence test. But this is a test for divergence. That a series does not diverge by this test doesn't necessarily mean that it converges. It might diverge by another test, and this one does.

Look at the partial sum of N terms, S_N, below:

Notice that that term-by-term, the element on the left sum is greater than the element on the right.

Now we can figure out (below) what the sum on the right is, and it's not zero, so this partial sum does not converge to zero, therefore neither does the series.

When a series diverges by the divergence test, it is a divergent series. When the divergence test is negative, as in this case, the series may converge, or it may diverge by another test, just as this one did. Just because the divergence test fails, does not mean that a series converges.

Does the series $\sum_{n = 1}^{\infty} \frac{2 n^2 + n^3}{4n^3 - 3n}$ converge?

Solution: The divergence test asks for the limit of the terms of the series as n → ∞. We'll use L'Hopital's rule:

$$\lim_{n\to\infty} \frac{2 n^2 + n^3}{4n^3 - 3n} = \lim_{n\to\infty} \frac{4n + 3n^2}{12n^2 - 3}$$

and another couple of rounds (often needed with L'Hopital's rule) gives the limit:

$$= \lim_{n\to\infty} \frac{4 + 6n}{24n} = \frac{1}{4}$$

Because the limit of the terms is not zero, we can safely conclude that this series does not converge. This is where the divergence test is really useful. We can move on from this series for most uses because it is divergent.