The Divergence Theorem for Series

We will now look at a fundamentally critical theorem that tells us that if a series is convergent then the sequence of terms $\{ a_n \}$ is convergent to 0, and that if the sequence of terms $\{ a_n \}$ does not diverge to $0$, then the series is divergent.

Theorem 1 (The Divergence Theorem for Series): If the series $\sum_{n=1}^{\infty} a_n$ is convergent, then $\lim_{n \to \infty} a_n = 0$. If the $\lim_{n \to \infty} a_n ≠ 0$ or this limit does not exist, then the series $\sum_{n=1}^{\infty} a_n$ is divergent.

• Proof of Theorem 1: Consider the $n^{\mathrm{th}}$ partial sum $s_n = a_1 + a_2 + ... + a_{n-1} + a_n$ and the $(n-1)^{\mathrm{th}}$ partial sum $s_{n-1} = a_1 + a_2 + ... + a_{n-1}$. We note that $a_n = s_n - s_{n-1}$.

• Now if the series $\sum_{n=1}^{\infty} a_n$ converges then $\lim_{n \to \infty} s_n = S$ for some $S$. Similarly, $\lim_{n \to \infty} s_{n-1} = S$. Therefore $\lim_{n \to \infty} \left ( s_n - s_{n-1} \right ) = \lim_{n \to \infty} a_n = S - S = 0$. $\blacksquare$

It is important to recognize that the converse of theorem 1 which says if $\lim_{n \to \infty} a_n = 0$ then $\sum_{n=1}^{\infty} a_n$ is convergent is not true. Now with this in hand, we will look at an important corollary from theorem 1.

Corollary 1: If $\{ a_n \}$ is unbounded and $\lim_{n \to \infty} a_n = L$ where $L \neq 0$, then if $L > 0$ the series $\sum_{n=1}^{\infty} a_n$ diverges to $\infty$ and if $L < 0$ the series $\sum_{n=1}^{\infty} a_n$ diverges to $-\infty$.

Divergence Test for Series

The Divergence Theorem is critically important as it provides us with a test to see whether a series is divergent. The test is as follows given some series $\sum_{n=1}^{\infty} a_n$.

• If $\lim_{n \to \infty} a_n \neq 0$, then the series $\sum_{n=1}^{\infty} a_n$ is divergent by the divergence theorem.

• If $\lim_{n \to \infty} a_n = 0$, then nothing can be said about the series $\sum_{n=1}^{\infty} a_n$, that is the series $\sum_{n=1}^{\infty} a_n$ may be convergent or divergent.

We will now look at some examples of applying the divergence test.

Example 1

Can we tell if the series $\sum_{n=1}^{\infty} n^2 - 1$ is divergent?

We will first take the limit as $n \to \infty$ of the sequence of terms $\{ a_n \}$ corresponding to this series.

(1)

\begin{align} \lim_{n \to \infty} n^2 - 1 = \infty \end{align}

Therefore since $\lim_{n \to \infty} n^2 - 1 \neq 0$, it follows that the series $\sum_{n=1}^{\infty} n^2 - 1$ is divergent by the divergence theorem.

Example 2

Can we tell if the series $\sum_{n=1}^{\infty} \frac{n^2 + n + 1}{2n^3 - 1}$ is divergent?

Like the last example, we will take the limit as $n \to \infty$ of the corresponding sequence of terms for this series:

(2)

\begin{align} \lim_{n \to \infty} \frac{n^2 + n + 1}{2n^3 - 1} \\ = \lim_{n \to \infty} \frac{\frac{1}{n} + \frac{1}{n^2} + \frac{1}{n^3}}{2 - \frac{1}{n^3}} = 0 \end{align}

Since $\lim_{n \to \infty} a_n = 0$, we conclude that the divergence theorem fails to tell us if the series is convergent or divergent.

Example 3

Can we tell if the series $\sum_{n=1}^{\infty} \sin^2 n+ \cos^2 n$ is divergent?

Recall that $\sin ^2 n + \cos ^2 n = 1$. Therefore $\sum_{n=1}^{\infty} \sin^2 n+ \cos^2 n = \sum_{n=1}^{\infty} 1$. Now we note that $\lim_{n \to \infty} 1 = 1 \neq 0$ and so this series diverges by the divergence theorem.

Example 4

Can we tell if the series $\sum_{n=1}^{\infty} \frac{(n^2)!}{n!}$ is divergent?

We note that $a_n = \frac{(2n)!}{n!} = \frac{1 \cdot 2 \cdot ... \cdot n \cdot (n+1) \cdot ... \cdot n^2}{1 \cdot 2 \cdot ... \cdot n} = (n+1)(n+2)...(2n)$

Therefore $\lim_{n \to \infty} a_n = \lim_{n \to \infty} (n+1)(n+2)...(n^2) = \infty$. By the divergence theorem since $\lim_{n \to \infty} \neq 0$ we conclude that the series $\sum_{n=1}^{\infty} \frac{(n^2)!}{n!}$ is divergent.