Convergence of the series: $\sum_{n\geq1}\frac{(-1)^n\arctan (n)}{n+n^{1/2}}$ $$\sum_{n=1}^{\infty} (-1)^n \tan^{-1}(n)/(n+(n)^{1/2}).$$ 
I know that the series is not absolutely converges.
I want to prove using Alternative test. I don't know how to prove that  sequence  $ tan^{-1}n/(n+(n)^{1/2})$ is decreasing sequence.
 A: For large $n$ we have

$$ \arctan(n) \sim \frac{\pi}{2}-\frac{1}{n}.  $$

A: You may write
$$
\sum_{n\geq1}\frac{(-1)^n\arctan (n)}{n+n^{1/2}}=\frac{\pi}{2}\sum_{n\geq1}\frac{(-1)^n}{n+n^{1/2}}-\sum_{n\geq1}\frac{(-1)^n}{n+n^{1/2}}\arctan \left(\frac1n\right).
$$ The first series on the right hand side is convergent by the alternate series test, the second series is absolutely convergent, since as $n \to \infty$,
$$
\left|\frac{(-1)^n}{n+n^{1/2}}\arctan \left(\frac1n\right)\right|\leq
\frac{1}{n+n^{1/2}}\arctan \left(\frac1n\right) \sim \frac{1}{n^{2}}.
$$ Your initial series is then convergent.
A: Certainly using a comparison test is easier, but here's what the Alternating Series test requires.
1)  Since the arctangent function has an upper bound of $ \ \frac{\pi}{2} \ $ , it is reasonably clear that $$ \ \lim_{n \rightarrow \infty} \ \ \frac{\arctan \ n}{n \ + \ n^{1/2}} \ \ = \ \ 0 \ \ , $$
so the series passes the "test for divergence".
2) Differentiating the real function $  \ \frac{\arctan \ x}{x \ + \ x^{1/2}} \ $  gives us
$$ \  \frac{(x \ + \ \sqrt{x}) \ \cdot \ \frac{1}{1 \ + \ x^2} \ - \ \arctan \ x \ \cdot \ ( 1 \ + \ \frac{1}{2 \ \sqrt{x}})}{( \ x \ + \ \sqrt{x} \ )^2}  $$
$$ = \ \ \frac{(x^2 \ + \ x\sqrt{x})  \ - \ \arctan \ x \ \cdot \ ( x \ + \ \frac{1}{2} \ \sqrt{x}) \ (1 \ + \ x^2)}{x \ (1 \ + \ x^2) \ ( \ x \ + \ \sqrt{x} \ )^2}  \ \ . $$
As  $ \ x \ \rightarrow \ \infty\ $ , the denominator grows as $ \ x^5 \ $  , the first term in the numerator as $ \ x^2 \ $ , and the second term as $ \ x^3 \ $ .  Since the arctangent factor only tends to a constant, this derivative will tend to roughly $ \ \frac{1 \ - \ \frac{\pi}{2} \ x}{x^3} \ $ .  So at least beyond some finite value of  $ \ n \ $ , the terms are always decreasing.  
So the conditions for alternating-series convergence appear to be satisfied.
