Calculus: Early Transcendentals 8th Edition

Since there is a factor $(-1)^{n-1}$ in the summation, it is an Alternating Series. We can use the Alternating Series Test by checking that the two following conditions are true: 1. $|a_{n}|\geq |a_{n+1}|$ for all $n$ 2.$\lim\limits_{n \to \infty}|a_{n}|=0$ The first condition means that the absolute value of the terms have to get smaller, and the second condition tells us how small should they get: $0!$ Let $a_{n}=(-1)^{n-1}\frac{\sqrt n}{n+1}$, then $|a_{n}|= \frac{\sqrt n}{n+1}$, (1) $|a_{1}|=0.5,|a_{2}|=0.4714,|a_{3}|=0.4330,....$ As we can see, the absolute value of the terms are indeed getting smaller. (2) $\lim\limits_{n \to \infty}|a_{n}|=\lim\limits_{n \to \infty}\frac{\sqrt n}{n+1}$ $=\lim\limits_{n \to \infty}\frac{\sqrt n}{n}$ $=\lim\limits_{n \to \infty}\frac{1}{\sqrt n}$ $=\frac{1}{\infty}$ $=0$ Thus, both conditions are verified. Hence, the series is convergent.