<s>[INST] Whether the given series are convergent or divergent: Whether the given series is absolutely convergent, conditionally convergent or divergent:
$a)\displaystyle\sum_{n=1}^{\infty}(-1)^n\frac{\cos nx}{n^2}:|(-1)^n\frac{\cos nx}{n^2}|\le\dfrac{1}{n^2}\implies$ Absolutely convergent.
$b)\displaystyle\sum_{n=1}^{\infty}(-1)^n\dfrac{n}{n+2}:(-1)^n\dfrac{n}{n+2}\to0\implies\dfrac{n}{n+2}\to0!\implies$ Divergent.
$c)\displaystyle\sum_{n=1}^{\infty}(-1)^{n-1}\dfrac{1}{2n+3}:u_n=\dfrac{1}{2n+3}\\\implies\dfrac{u_n}{u_{n+1}}=\dfrac{2n+3}{2n+5}=\dfrac{1+\dfrac{3}{2n}}{1+\dfrac{5}{2n}}=({1+\dfrac{3}{2n}})(1+\dfrac{5}{2n})^{-1}=({1+\dfrac{3}{2n}})(1-\dfrac{5}{2n}+\cdots)=1-\dfrac{1}{n}+O(\dfrac{1}{n^2})\\\implies \text{Not absolutely convergent. So conditionally convergent by Leibnitz test.}$ 
Am i right?
 [/INST] For (b), $\frac{n}{n+2} \to 1$,
so the series is absolutely
fairly violently divergent.
The sum of consecutive even-odd terms is
$\frac{2n}{2n+2} - \frac{2n+1}{2n+3}
= \frac{2n(2n+3)-(2n+2)(2n+1)}{(2n+2)(2n+3)}
= \frac{4n^2+6n-(4n^2+3n+2)}{(2n+2)(2n+3)}
= \frac{3n-2}{(2n+2)(2n+3)}
$,
and the sum of consecutive odd-even terms is
$-\frac{2n+1}{2n+3}+\frac{2n+2}{2n+4} - 
= \frac{-(2n+1)(2n+4)+(2n+2)(2n+3)-}{(2n+3)(2n+4)}
= \frac{-(4n^2+7n+4)+(4n^2+6n+6)}{(2n+3)(2n+4)}
= \frac{-n+2}{(2n+3)(2n+4)}
$,
so the series is not even conditionally convergent.
</s>