20 - Alternating Series Test

Alternating Series Test

Let $(u_k)$ be a real sequence, and consider the series

$$\sum _{k=1}^{\infty }(-1{)}^{k-1}{u}_k$$

If

1. $u_k\ge 0$ for $k\ge 1$
2. $(u_k)$ is decreasing, that is $u_{k+1}\le u_k$ for $k\ge 1$
3. $u_k\to 0$ as $k\to \infty$
   Then

$$\sum _{k=1}^{\infty }(-1{)}^{k-1}{u}_k\text{ converges}$$

Proof

Let $s_n={\sum }_{k=1}^n(-1{)}^{k-1}{u}_k$

1. $(s_{2n})$ is bounded above as

$$s_{2n}=u_1-(u_2-u_3)-(u_4-u_5)-\cdots -(u_{2n-2}-u_{2n-1})-u_{2n}\le u_1$$

So $u_1$ is an upper bound of $(s_{2n})$
2) $(s_{2n})$ is increasing as

$$s_{2n+2}-s_{2n}=u_{2n+1}-u_{2n+2}\ge 0$$

So by the Monotone Sequences Theorem, $(s_{2n})$ converges
Suppose $s_{2n}\to s$ as $n\to \infty$
As

$$s_{2n+1}=s_{2n}+u_{2n+1}\to s+0=s\text{ as }n\to \infty \text{ by AOL}$$

So $(s_{2n+1})$ also converges to $s$

There is $N_1\in \mathbb{N}$ such that for $n\ge N_1$ then

$$|s_{2n}-s|<ϵ$$

There is $N_2\in \mathbb{N}$ such that for $n\ge N_2$ then

$$|s_{2n+1}-s|<ϵ$$

Let $N=\max \{2N_1,2N_2+1\}$ then for $n\ge N$ then

1. If $n$ is even then $n=2k$ for some $k$ so $|s_n-s|=|s_{2k}-s|<ϵ$
2. If $n$ is odd then $n=2k+1$ for some $k$ so $|s_n-s|=|s_{2k+1}-s|<ϵ$

Hence $|s_n-s|<ϵ$ so $(s_n)$ converges