📂Calculus

Alternating Series Test

Summary¹

The alternating series satisfying the following conditions $\sum\limits_{n = 1}^{\infty} (-1)^{n-1}b_{n}$ $(b_{n} \gt 0)$ converges.

1. $b_{n+1} \le b_{n} \quad \forall n$.
2. $\lim\limits_{n \to \infty} b_{n} = 0$.

Proof

First, consider the partial sum up to the even terms.

$$\begin{align*} s_{2} &= b_{1} - b_{2} \ge 0 \\ s_{4} &= s_{2} + (b_{3} - b_{4}) \ge s_{2} \\ s_{6} &= s_{4} + (b_{5} - b_{6}) \ge s_{4} \\ &\quad \vdots \\ s_{2n} &= s_{2n-2} + (b_{2n-1} - b_{2n}) \ge s_{2n-2} \end{align*}$$

Therefore, the following holds.

$$0 \le s_{2} \le s_{4} \le \cdots \le s_{2n} \le \cdots$$

Meanwhile, $s_{2n}$ can be written as follows.

$$s_{2n} = b_{1} - (b_{2} - b_{3}) - (b_{4} - b_{5}) - \cdots - (b_{2n-2} - b_{2n-1}) - b_{2n}$$

Since all terms inside the parentheses and $b_{2n}$ are positive, $s_{2n} \lt b_{1}$ holds for all $n$. That is, it is bounded above. Given that a monotonic and bounded sequence converges, $s_{2n}$ converges.

$$\lim\limits_{n \to \infty} s_{2n} = s$$

Now, considering the partial sum up to the odd terms, it can be seen that it converges to $s$ as follows.

$$\begin{align*} \lim\limits_{n \to \infty} s_{2n + 1} &= \lim\limits_{n \to \infty} s_{2n} + b_{2n + 1} \\ &= \lim\limits_{n \to \infty} s_{2n} + \lim\limits_{n \to \infty} b_{2n + 1} \\ &= s + 0 \\ &= s \end{align*}$$

Lemma

If $\lim\limits_{n \to \infty} a_{2n} = L$ and $\lim\limits_{n \to \infty} a_{2n+1} = L$, then $\lim\limits_{n \to \infty} a_{n} = L$.

By the lemma above, it is $\sum\limits_{n = 1}^{\infty} s_{n} = s$. That is, the series $\sum\limits_{n = 1}^{\infty} (-1)^{n-1}b_{n}$ converges.

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