Check whether the given series is conditionally convergent or absolutely convergent or divergent?

Question

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?

Answer 1

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.

Answer 2

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?