We start with the alternating series test. We need to check two conditions:
$$ \lim\limits_{n\to \infty} a_n=0 \hskip{2em} a_{n+1} \leq a_n \text{ for all } n$$
Checking the first condition:
$$ \lim\limits_{n\to\infty} \frac{1}{1+\sqrt{n}} \checkmark$$
As \(n\) approaches infinity, the denominator increases, and the value of the fraction approaches 0.
We can check the second condition by taking the derivative of the expression and confirming if the value is negative for all values. We can also write out the terms and check using inspection:
$$ \sum_{n=1}^\infty \frac{1}{1+\sqrt{n}} = \frac{1}{1+1} + \frac{1}{1+\sqrt{2}} + \frac{1}{1+\sqrt{3}} + \frac{1}{1+2} + \frac{1}{1+\sqrt{5}}...$$
Since the terms are decreasing, the second condition is met. The series converges. To check if the series converges absolutely, we need to check the convergence of the series:
$$ \Big| \sum_{n=1}^\infty \frac{(-1)^n}{1+\sqrt{n}} \Big | = \sum_{n=1}^\infty \frac{1}{1+\sqrt{n}} $$
We can perform a limit comparison test with the divergent \(p\)-series \(\dfrac{1}{\sqrt{n}}\) where \(p\leq 1\).
$$ \lim\limits_{n\to\infty} \frac{1/(1+\sqrt{n})}{1/\sqrt{n}} $$
$$ = \lim\limits_{n\to\infty} \frac{\sqrt{n}}{1+\sqrt{n}} = 1 $$
By the limit comparison test, both series diverge.
Since \(\displaystyle \sum_{n=1}^\infty \frac{(-1)^n}{1+\sqrt{n}} \) converges but \(\displaystyle \sum_{n=1}^\infty \frac{1}{1+\sqrt{n}} \) diverges, the original series diverges conditionally.