Divergence Test for Series

The simplest divergence test, called the Divergence Test, is used to determine whether the sum of a series diverges based on the series's end-behavior. It cannot be used alone to determine wheter the sum of a series converges.

Allow a series $n$ that has infinitely many elements. Even though it's infinitely long, the series having a nonzero "last term" would make the sum diverge. For example, the sum of the series $n = \{1, 1, 1, 1,...\}$ diverges, because it's always going to add $1$.

If $\lim_{k\to\infty}n_k \ne 0$ then the sum of the series diverges. Otherwise, the test is inconclusive.

We now know what an infinite series is, what it means for a series to diverge or converge, and how to test geometric series for convergence (we can even calculate their limit if they do converge!) We now want to build a toolkit which will allow us to determine whether a much wider range of series converge or diverge.

Let's start by demonstrating a simple test which can detect certain divergent series. Suppose we have a series $\sum_{n=1}^{\infty} a_{n}$ where the sequence $a_{n}$ converges to a nonzero limit $L$. For example, think of the constant sequence given by $a_{n} = 2$. The partial sums of this series are $s_{n} = 2n$, which is evidently unbounded, so the series diverges. More generally, any constant sequence $a_{n} = a$, where $a \neq 0$, will diverge.

Even more generally, we can show that if the series $\sum a_{n}$ converges, then the sequence $a_{n}$ converges to $0$. Or said differently, if the sequence $a_{n}$ does not converge to $0$, then $\sum a_{n}$ diverges. That is, the following theorem is true:

Theorem. (The nth term test for divergence.) Suppose the sequence $a_{n}$ does not converge to $0$. Then the series $\sum_{n=1}^{\infty} a_{n}$ diverges.

For a proof of this theorem, please see the end of this section. Notice that this test is applicable not just to situations where the sequence $a_{n}$ converges to a non-zero limit. It is also applicable when $a_{n}$ diverges to $\pm \infty$ or when $a_{n}$ simply diverges. Let's look at some examples.

Notice that in each of these examples, we used the nth term test to show that a series diverges. We never used the test to show that a series converges. This is not a coincidence! The nth term test, by itself, is insufficient to show that a series converges. Indeed, there are examples of sequences $a_{n}$ which converge to 0 yet whose series $\sum a_{n}$ diverge. For example,

$$\sum_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \ldots$$

diverges, yet its nth term converges to $0$. This series is so special that it is called the harmonic series. In a future section we will see a simple way to prove the divergence of the harmonic series; however, there is actually an elementary way to show that this series diverges.

The clever trick is to find lower bounds for the partial sums of this serieswhich diverge to $\infty$. In particular, let's group various pieces of this series as follows: $$\sum_{n=1}^{\infty} \left( \frac{1}{1} \right) + \left( \frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right) + \left( \frac{1}{5} + \ldots + \frac{1}{13} \right) + \ldots$$What we're doing here is grouping terms of this sums into groups of size $3^{n}$ (above we have groups of size 1, 3, and 9, the grouping continues to sums of size 27, 81, etc.). It is easy to show that each group has sum greater than 1. But then this means that the partial sums of the harmonic series must diverge to $\infty$.

This example shows that if a sequence $\sum a_{n} \rightarrow 0$, the nth term test by itself cannot say anything about whether $\sum a_{n}$ diverges or converges. As a matter of fact, the majority of the remaining tests you will learn about are precisely to help us answer the question of what happens when $\sum a_{n} \rightarrow 0$.

Bonus section. If you're interested in a proof of the nth term test for divergence, read here. (TODO: Fill this in later!!!)

Summand Test for Series Convergence (Divergence Test)

If an infinite series converges, then the individual terms (of the underlying sequence being summed) must converge to $0$. This can be phrased as a simple divergence test: If $\lim_{n\to\infty}a_n$ either does not exist, or exists but is nonzero, then the infinite series $\sum_n a_n$ diverges. Otherwise, the test is inconclusive.