*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.

Note that I've kept the an term on the outside to give

So we have the equation

which rearranges to

Now we have an expression for **a _{n}** of which we can take advantage. Here is the substance of the proof:

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

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:

The limit is not zero, so the series diverges. Notice that for large enough n, the terms are approximately 1/2, so the sum would have a couple of smaller terms, then be a series of infinitely-many 1/2's added together, like this:

The graph shows that the size of the terms is relatively constant as they approach the limit of 1/2, and the resulting sum grows without bound. Clearly the series diverges, and that's all we need to conclude if we're looking for convergence.

Here are the first few terms of this series:

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

So it looks like this series converges: The limit of the terms is zero. 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 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, the series may converge, or it may diverge by another test. Just because the divergence test fails, does not mean that a series converges.

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

and another couple of rounds gives the limit:

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.

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