Alternating Series Test

Theorem

Let $\left \langle {a_n} \right \rangle$ be a decreasing sequence of positive terms in $\R$ which converges with a limit of zero.

That is, let $\forall n \in \N: a_n \ge 0, a_{n+1} \le a_n, a_n \to 0$ as $n \to \infty$.

Then the series $\displaystyle \sum_{n=1}^\infty \left({-1}\right)^{n-1} a_n = a_1 - a_2 + a_3 - a_4 + \ldots$ converges.

Proof

First we show that for each $n > m$, we have $0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_n \le a_{m+1}$.

This will be achieved by means of the Second Principle of Mathematical Induction.

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition $0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_n \le a_{m+1}$.

It certainly holds for $n = 1$ as $a_{m+1} \ge 0$ by definition and $a_{m+1} \le a_{m+1}$.

It also holds for $n = 2$ as $a_{m+2} \le a_{m+1}$ and so $0 \le a_{m+1} - a_{m+2} \le a_{m+1}$.

These are the base cases.

Let $b_k = a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_{k}$ (this simplifies the algebra). The $\pm$ signifies the fact that $a_k$ is positive for $k$ odd and negative for $k$ even.

• Suppose that $\forall j \le k: P \left({j}\right)$ holds, that is, that $0 \le b_j \le a_{m+1}$.

This is our induction hypothesis.

We now show that $\forall k: P \left({k}\right) \implies P \left({k+1}\right)$. This is the induction step.

• Suppose $k$ is odd.

We have by induction hypothesis that $0 \le b_k \le a_{m+1}$.

Because $P \left({k}\right)$ holds, $0 \le b_{k-1} + a_k \le a_{m+1}$.

But as $a_k \ge a_{k+1}$, $a_k - a_{k+1} \ge 0$ and so $0 \le b_{k-1} + \left({a_k - a_{k+1}}\right) = b_{k+1}$.

But as $b_k \le a_{m+1}$ it follows that $b_k - a_{k+1} = b_{k+1} \le a_{m+1}$.

So $0 \le b_{k+1} \le a_{m+1}$, or $0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots - a_{k+1} \le a_{m+1}$.

Thus for odd $k$ it follows that $P \left({k}\right) \implies P \left({k+1}\right)$.

• Now suppose $k$ is even.

We have by induction hypothesis that $0 \le b_k \le a_{m+1}$.

Then $0 \le b_k + a_{k+1} = b_{k+1}$.

Because $P \left({k}\right)$ holds, $0 \le b_{k-1} - a_k \le a_{m+1}$.

But as $a_k \ge a_{k+1}$, $a_k - a_{k+1} \ge 0$ and so $b_{k+1} = b_{k-1} - a_k + a_{k+1} = b_{k-1} - \left({a_k - a_{k+1}}\right) = b_{k+1} \le a_{m+1}$.

So $0 \le b_{k+1} \le a_{m+1}$, or $0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots + a_{k+1} \le a_{m+1}$.

Thus for even $k$ it follows that $P \left({k}\right) \implies P \left({k+1}\right)$.

So for both even and odd $k$ it follows that $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Second Principle of Mathematical Induction.

Therefore for each $n > m$, we have $0 \le a_{m+1} - a_{m+2} + a_{m+3} - \cdots \pm a_n \le a_{m+1}$.

• Now, let $\left \langle {s_n} \right \rangle$ be the sequence of partial sums of the series $\displaystyle \sum_{n=1}^\infty \left({-1}\right)^{n-1} a_n$.

Let $\epsilon > 0$.

Since $a_n \to 0$ as $n \to \infty$, it follows that $\exists N: \forall n > N: a_n < \epsilon$.

But $\forall n > m > N$, we have:

	\(\displaystyle \)	\(\displaystyle \mid {s_n - s_m}\mid\)	\(=\)	\(\displaystyle \mid {\left({a_1 - a_2 + a_3 - \ldots \pm a_n}\right) - \left({a_1 - a_2 + a_3 - \ldots \pm a_m}\right)}\mid\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \mid {\left({a_{m+1} - a_{m+2} + a_{m+3} - \ldots \pm a_n}\right)}\mid\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\le\)	\(\displaystyle a_{m+1}\)	\(\displaystyle \)	from the above
	\(\displaystyle \)	\(\displaystyle \)	\(<\)	\(\displaystyle \epsilon\)	\(\displaystyle \)	as $m+1 > N$

Thus we have shown that $\left \langle {s_n} \right \rangle$ is a Cauchy sequence.

The result follows from the fact that a Cauchy sequence is convergent.

$\blacksquare$

Sources

• K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $\S 6.13$