<s>[INST] Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent Show if $\sum_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$ is absolute convergent or divergent
First i subbed numbers in
$$\lim_{n \to \infty} \frac{(-1)^n}{1+\sqrt{n}} = \frac{-1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} - \frac{-1}{1+\sqrt{3}}$$ 
So it's divergent
1 LIMIT
$$\lim_{n \to \infty} \frac{1}{1+\sqrt{n}}=0 \quad \text{hence divergent} $$
2 $a_{n}$ and $a_{n+1}$
$$ \frac{1}{1+\sqrt{n}}>\frac{1}{1+\sqrt{n+1}} $$
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}}> 0 $$
hence increasing
or 
$$ \frac{1}{1+\sqrt{n}}-\frac{1}{1+\sqrt{n+1}} =  \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n+1}\sqrt{n+1}} $$
$$ \therefore \sqrt{n+1}-\sqrt{n}> 0$$ hence decreasing
$ \sum_{n=1}^{\infty} \left\lvert \frac{-1^n}{1+\sqrt{n}}  \right\rvert = \sum_{n=1}^{\infty}   \frac{1}{1+\sqrt{n}}   $
$$ \sum_{n=1}^{\infty} \frac{1}{1+\sqrt{n}} = \frac{1}{1+\sqrt{1}} + \frac{1}{1+\sqrt{2}} + \frac{1}{1+\sqrt{3}} $$
if this is convergent then the whole thing is absolutely convergent but i don't know how to prove this

UPDATE
I got something out for the second part

 [/INST] There are so many mistakes (and so severe) in your "solution" that it is a superhuman task to discuss them. Your series is convergent, Leibniz's test solves the problem in a second. The series is not absolutely convergent by an application of the limit comparison test (compare your series with $\sum \frac 1 {\sqrt n}$ which is known to be divergent).
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