Alternating Series Test

Theorem

• Given ${\sum }_{n=i}^{\infty }(-1{)}^{n+1}{b}_n$
• With $b_n>0$
• If $b_n\ge b_{n+1}$ $\forall n\in \mathbb{N}$, (just decreasing, not necessarily Strictly Decreasing)
• If ${\lim }_{n\to \infty }{b}_n=0$
• Then, ${\sum }_{n=i}^{\infty }(-1{)}^{n+1}{b}_n$ converges

Proof

1. Given ${\sum }_{n=i}^{\infty }(-1{)}^{n+1}{b}_n$, $b_n>0$
2. Suppose $b_n>b_{n+1},\forall n\in \mathbb{N}$
3. Suppose ${\lim }_{n\to \infty }{b}_n=0$
4. We want to show, ${\sum }_{n=i}^{\infty }(-1{)}^{n+1}{b}_n$ converges. We prove by definition
5. Consider odd-index set of $\{S_n\}$. This is a Monotonic Sequence
   1. Then, by BMCT, this set converges
6. Consider even-index set of $\{S_n\}$. This is a Monotonic Sequence
   1. Then, by BMCT, this set converges
7. We find later that ${\lim }_{n\to \infty }{S}_{2n}={\lim }_{n\to \infty }{S}_{2n+1}=s$, then by definition, set converges
8. $\square$

Example

Prove ${\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{\ln (n)}$ converges

• Note $b_n=\frac{1}{\ln (n)}>0$
• Note that $\ln (n)<\ln (n+1)$ as ln is increasing
• $⟹\frac{1}{\ln (n)}>\frac{1}{\ln (n+1)}$
• Thus, $b_n$ is decreasing
• Consider ${\lim }_{n\to \infty }{b}_n=0$
• Then, by AST, ${\sum }_{n=1}^{\infty }\frac{(-1{)}^n}{\ln (n)}$ converges