Answer
Convergent
Work Step by Step
Since, there is a factor $(-1)^{n-1}$ in the summation, it is an Alternating Series. We can use Alternating Series Test by checking 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 the absolute value of the terms have to get smaller, and the second condition tells 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 that the absolute value of the term are indeed getting small.
(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.
