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?

Note that the ratio test actually has absolute value bars:
$$\lim_{n\to\infty}\left|\frac{a_{n+1}}{a_n}\right|$$
You should instead use the alternating test:
$$\lim_{n\to\infty}a_n=0$$
thus, it converges.  To see it does not converge absolutely, note that
$$\frac1{2n+3}>\frac1{3n}$$
For the last two:
ii) Use the term test.
iii) Check for absolute convergence with the ratio test.
