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

 A: 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).
A: First, you have some errors I think in your writeup. $-1^n$ likely should have parentheses around the $-1$.  It does change the convergence or divergence of this particular series. There is a different treatment for alternating series. But I'll proceed to answer the problem as written. 
Second, your first use of $a_n$ and $a_{n+1}$ ends with a conclusion that the sequence of terms is increasing. But when fractions get bigger denominators, they get smaller... So this is a decreasing sequence. 
Third, the series diverges by comparison with the $p-$series. If you are not familiar with this, you should look it up. It's a standard thing in most textbooks. 
Now your series is not exactly a $p-$series, but it behaves like one. To show that it diverges, you can take
$$\lim_{n\to\infty} \dfrac{1/\sqrt n} {1/(1+\sqrt n)} $$
to see why. Since this limit is finite and positive, both series behave the same way. 
Now if your series is alternating, i.e. the general term contains $(-1)^n,$ then the series converges conditionally.  This is because the sequence of the absolute value of terms strictly decreases to zero, while the series with the absolute value of terms diverges (as we mentioned above). 
Either way, your series does not converge absolutely! For this to occur, the alternating series and its associated series with the absolute values of terms must BOTH converge. 
A: Let $a_n=\frac{(-1)^n}{1+\sqrt{n}}$.
$$ a_{2n}+a_{2n+1} = \frac{1}{1+\sqrt{2n}}-\frac{1}{1+\sqrt{2n+1}} =  \frac{\sqrt{2n+1}-\sqrt{2n}}{(1+\sqrt{2n})(1+\sqrt{2n+1})}  =  \sqrt{2n} \frac{\sqrt{1+\frac{1}{2n}}-1}{(1+\sqrt{2n})(1+\sqrt{2n+1})} \sim \frac{n^{-\frac{3}{2}}}{4\sqrt{2} } = O\left(\frac{1}{n^{\frac{3}{2}}}\right)$$
Therefore the serie $\sum (a_{2n}+a_{2n+1})$ is convergent and so is the serie $\sum a_n$.
Moreover, $|a_n| \sim \frac{1}{\sqrt{n}} $ so your serie is not absolutely convergent.
A: $$\sum_{n=1}^{\infty}\left|\frac{(-1)^n}{1+\sqrt{n}}\right|=\sum_{n=1}^{\infty} \frac1{1+\sqrt{n}}>\frac12\sum_{n=1}^{\infty} \frac1{\sqrt{n}}$$ is divergent.
A: Absolute convergence

If
  $$ \sum \left|a_n\right| =\mbox{convergent}$$
  Then
  $$ \sum a_n =\mbox{absolutely convergent}$$
  Note that an absolutely convergent series is also convergent.

Conditional convergence

If
  $$ \sum \left|a_n\right| =\mbox{divergent}$$
  And
  $$ \sum a_n =\mbox{convergent}$$
  Then
  $$ \sum a_n =\mbox{conditionally convergent}$$

So now 
$$\sum\limits_{n=1}^{\infty} \left|\frac{(-1)^n}{1+\sqrt{n}}\right|=\sum\limits_{n=1}^{\infty} \left|\frac{1}{1+\sqrt{n}}\right|$$
Since
$$ \sum\limits_{n=1}^{\infty}\left|\frac{1}{\sqrt{n}+\sqrt{n}}\right|\leq \sum\limits_{n=1}^{\infty}\left|\frac{1}{1+\sqrt{n}}\right|$$
And
$$ \sum\limits_{n=1}^{\infty}\left|\frac{1}{\sqrt{n}+\sqrt{n}}\right| =\frac12\sum\limits_{n=1}^{\infty}\left|\frac{1}{\sqrt{n}}\right|=\mbox{divergent}$$
Then by the comparison test
$$\sum\limits_{n=1}^{\infty}\left|\frac{1}{1+\sqrt{n}}\right|=\mbox{divergent}$$
Now for the following series
$$\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}}$$
Since
$$\lim\limits_{n\to\infty} \frac{1}{1+\sqrt{n}}=0$$
And
$$  \frac{1}{1+\sqrt{n}} \gt  \frac{1}{1+\sqrt{n+1}} $$
Then by the alternating series test
$$\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}} =\mbox{convergent}$$
Putting everything together, we see that
$$\sum\limits_{n=1}^{\infty} \frac{(-1)^n}{1+\sqrt{n}} =\mbox{conditionally convergent}$$
