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?