## Calculus 8th Edition

Alternating series test: Suppose that we have series $\Sigma a_n$, such that $a_{n}=(-1)^{n}b_n$ or $a_{n}=(-1)^{n+1}b_n$, where $b_n\geq 0$ for all $n$. Then if the following two condition are satisfied the series is convergent. 1. $\lim\limits_{n \to \infty}b_{n}=0$ 2. $b_{n}$ is a decreasing sequence. Given: $\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}-....$ General Term $a_{n}=(-1)^{n+1}\frac{2}{2n+1}$ Thus, $\frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\frac{2}{11}-....=\Sigma_{n=1}^{\infty}(-1)^{n+1}\frac{2}{2n+1}$ In the given problem, $b_{n}=\frac{2}{2n+1}$ which satisfies both conditions of Alternating Series Test as follows: 1. Let $f(x)=\frac{2}{2x+1}$ and $f'(x)=\frac{-2.2}{(2x+1)^{2}} \lt 0$ Thus, $f(x)$ is deceasing , so $b_{n}=\frac{2}{2n+1}$ is decreasing because the denominator is increasing. 2. $\lim\limits_{n \to \infty}b_{n}=\lim\limits_{n \to \infty}\frac{2}{2n+1}=\frac{2}{\infty}=0$ Hence, the given series is convergent by Alternating Series Test.