Answer
$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ is conditionally convergent
Work Step by Step
We have, \[\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}\]
Compare $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ with $\displaystyle\sum_{n=1}^{\infty} (-1)^{n-1}a_n$
\[\Rightarrow a_n=\frac{1}{\sqrt{n}}\]
$a_n$ is positive term and monotonically decreasing,
and \[\lim_{n\rightarrow\infty}a_n=\lim_{n\rightarrow\infty}\frac{1}{\sqrt{n}}=0\]
By Alternating Series Test,
$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ is convergent.$\;\;\;\;\;\;\;\;\;\;\ldots(1)$
Compare $\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ with $\displaystyle\sum_{n=1}^{\infty} b_n$
\[\Rightarrow b_n=\frac{(-1)^{n-1}}{\sqrt{n}}\]
\[|b_n|=\left|\frac{(-1)^{n-1}}{\sqrt{n}}\right|=\frac{1}{\sqrt{n}}(=c_n)\]
Also $\displaystyle\sum_{n=1}^{\infty}c_n=\displaystyle\sum_{n=1}^{\infty}\frac{1}{\sqrt{n}}$ is divergent by P-Series Test.
$\Rightarrow \displaystyle\sum_{n=1}^{\infty}\left|\frac{(-1)^{n-1}}{\sqrt{n}}\right|$ is divergent.$\;\;\;\;\;\;\;\;\;\;\;\ldots (2)$
Using (1) and (2)
$\displaystyle\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{\sqrt{n}}$ is conditionally convergent.