Check whether the given series is conditionally convergent or absolutely convergent or divergent? Check whether the given series is conditionally convergent or absolutely convergent or divergent? 
(i)$\displaystyle\sum_{n=1}^\infty (-1)^n \frac 1 {2n+3}$
(ii)$\displaystyle\sum_{n=1}^\infty (-1)^n \frac n {n+2}$
(iii)$\displaystyle\sum_{n=1}^\infty (-1)^n \frac {n\log n} {e^n}$
MY TRY:(i)$\displaystyle\sum_{n=1}^\infty (-1)^n \frac 1 {2n+3}$ ,$\frac {a_{n+1}} {a_{n}}=-1<1$,so the series convergent.
But for $\displaystyle\sum_{n=1}^\infty  \frac 1 {2n+3}$, $\frac {a_{n+1}} {a_{n}}=1$. So how can we conclude anything for absolutely convergent?

Hints:
(i) $\;\frac1{2n+3}\;$ is monotone descending, so this is a Leibniz series. Without the absolute value though compare to the harmonic series
(ii) What is the limit of the series' sequence?
(iii) Use the ratio test without the $\;(-1)^n\;$ . What can you deduce from this?
