• Consider the associated sequence $\{s_n\}$ of partial sums, where $s_n=\sum _{k=1}^n a_k$.
• Try to find an explicit formula for the term $s_n$. If you can find such a formula, analyze $\lim _{n \to \infty s_n}$.
  • If the limit exists, $\sum _{k=k_0} a_k$ converges, and if we can determine that $\lim _{n \to \infty } s_n =L$, then $\sum _{k=k_0} a_k=L$.
  • If $\lim _{n \to \infty } s_n$ does not exist, then $\sum _{k=k_0} a_k$ diverges.
• If an explicit formula for $s_n$ cannot be found, further analysis is needed.

In the previous section, we studied two types of series where we could find an explicit formula for $s_n$, but unfortunately, this is not always easy or possible. Fortunately, it is not always necessary to do this in order to determine whether $\lim _{n \to \infty } s_n$ exists. Consider the example below.

Determine if the series $\sum _{k=1}^{\infty } \frac {k+2}{2k+1}$ converges or diverges.

Let’s think about what we are trying to do here heuristically. If we set $a_k = \frac {k+2}{2k+1}$, we can write out the first several terms in the sequence $\{a_n\}_{n=1}$.

$$\frac {3}{5}, \frac {4}{7}, \frac {5}{9}, \frac {6}{11},\frac {7}{13}, \ldots$$ We can observe also that $\lim _{n \to \infty } a_n = \answer [given]{\frac {1}{2}}$, so eventually, the terms in this list become as close to $\frac {1}{2}$ as we want. Conceptually, we can interpret that when trying to compute the series, we will have to add infinitely many numbers together that are very close to $\frac {1}{2}$. Such an attempt cannot produce a finite result, so we expect the series to diverge.

Of course, this is not a formal proof or an acceptable mathematical argument, but it is good intuition. In order to formalize the argument, recall that we have to set $s_n = \sum _{k=1}^n a_k$ and study whether $\lim _{n \to \infty } s_n$ exists. While we do not have an explicit formula for $s_n$, we do have a recursive formula, which we recall below.

Since $a_n = \frac {n+2}{2n+1}$, we can write

$$s_n = s_{n-1}+\frac {n+2}{2n+1}.$$

Now, if $\lim _{n \to \infty } s_n$ exists and is equal to $L$, we have that $\lim _{n \to \infty } s_{n-1}=L$ as well, so taking the limit of both sides of the above equation gives

$$\begin{align*} \lim_{n \to \infty} s_n &= \lim_{n \to \infty} s_{n-1}+ \lim_{n \to \infty} \frac{n+2}{2n+1} \\ L &= L +\frac{1}{2}. \end{align*}$$

This statement is blatantly false, so our underlying assumption that $\lim _{n \to \infty } s_n$ exists is false as well. Since $\lim _{n \to \infty } s_n$ therefore does not exist, $\sum _{k=1}^{\infty } \frac {k+2}{2k+1}$ must diverge.

As it turns out, the above argument can be used to make a very important observation; if $\{a_n\}$ is a sequence for which $\sum _{k=k_0}^{\infty } a_k$ converges, then $\lim _{n \to \infty } a_n =0$. This result is fundamentally important, so we capture it in a theorem.

The divergence test

Stated in plain English, the above test ensures that if the terms in a sequence do not tend to zero, then we cannot add all of the terms in that sequence together.

This test gives us a quick way to determine if some series diverge.

In the last example, perhaps the fact that the terms in $a_n = \sin (n)$ fluctuate in sign will ensure that the series cannot be infinite. To think about this, let’s turn to the sequence of partial sums. To gain a bit of visual perspective about what is happening, note that the $n$-th term in the sequence of partial sums here is

$$s_n = \sum _{k=1}^n \sin (k) = \sin (1)+\sin (2)+\ldots +\sin (n).$$

Plotting several such terms reveals that the terms sequence of partial sums $\{s_n\}$ seem to fluctuate.

While we will not show it here, the sequence $\{s_n\}$ is bounded; the reason that $\lim _{n \to \infty } s_n$ does not exist is due to the fact that the terms fluctuate (meaning that the sequence is never eventually monotonic).

Implications of the divergence test

Let’s summarize the important points from the previous discussion.

• If $\sum _{k=k_0}^\infty a_k$ converges, then $\lim _{n \to \infty } a_n =0$.
• If $\lim _{n \to \infty } a_n \neq 0$ (including the case where the limit does not exist), then $\sum _{k=k_0}^{\infty } a_k$ diverges.

While divergence test was straightforward to apply in the previous examples, there is a major point to address about what it does not say.

We’ve actually seen an example of this in action.

Said another way:

Restating this point again (because it is very important): passing the divergence test means that a series has a chance to converge. The divergence test cannot tell us whether a series converges.

There are many questions that require that you now have a firm grasp on the concepts presented thus far. We summarize the important points made thus far, then give many examples that require you to synthesize them.

• There are two fundamental questions we can ask of any sequence.
  • Do the numbers in the list approach a finite value?
  • Can I sum all of the numbers in the list and obtain a finite result?

  These questions can be asked of a given sequence $\{a_n\}$ and can also be asked about $\{s_n\}$ or any sequence constructed from it.

• Given a sequence $\{a_n\}_{n=n_0}$, we construct the sequence of partial sum $\{s_n\}_{n=n_0}$ whose $n$-th term is given by the formula $s_n = \sum _{k=k_0}^n a_k$.
  • The symbols $\sum _{k=k_0}^{\infty } a_k$ and $\lim _{n \to \infty } s_n$ are the same.
  • By definition $\sum _{k=k_0} a_k$ converges if $\lim _{n \to \infty } s_n$ exists and in this case, the value of each is the same.
  • By definition $\sum _{k=k_0} a_k$ diverges if $\lim _{n \to \infty } s_n$ does not exist (which includes if the limit is infinte).
• If the limit of a sequence is not zero, the sum of its terms diverges.
• If a series converges, the limit of the sequence whose terms is being summed is zero.
• If the limit of a sequence is zero, more information is needed to determine whether the sum of its terms converges or diverges.

To answer the following questions, make sure that you understand exactly what is given in the statement of the question first, then try to synthesize the material above.