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?

  • $\begingroup$ (i) You can't use the ratio test here as this is not a positive series. $\endgroup$ – DonAntonio Dec 26 '16 at 12:57
  • $\begingroup$ @DonAntonio I thought you could, but that you need absolute value bars. $\endgroup$ – Simply Beautiful Art Dec 26 '16 at 13:00
  • $\begingroup$ @SimpleArt Yes, thats right...but there is no absolute value in the work the OP wrote. $\endgroup$ – DonAntonio Dec 26 '16 at 13:01

Note that the ratio test actually has absolute value bars:


You should instead use the alternating test:


thus, it converges. To see it does not converge absolutely, note that


For the last two:

ii) Use the term test.

iii) Check for absolute convergence with the ratio test.

  • $\begingroup$ I understood for (i),(ii) but for (iii) For absolute convergence of $\displaystyle\sum_{n=1}^\infty (-1)^n \frac {n\log n} {e^n}$, $a_n=\frac {n\log n} {e^n}$.So $\displaystyle\lim _{n\to \infty}\left|\frac {a_{n+1}} {a_{n}}\right|=\displaystyle\lim _{n\to \infty}\left|\frac 1 e \frac {(n+1)\log (n+1)} {n\log n}\right|$....then? $\endgroup$ – MatheMagic Dec 26 '16 at 13:25
  • $\begingroup$ Split it as follows:$$\frac{n+1}n\frac{\log(n+1)}{\log(n)}$$Take L'Hospital's rule on each piece separately. @SubratKumarJena $\endgroup$ – Simply Beautiful Art Dec 26 '16 at 13:27
  • $\begingroup$ $\displaystyle\lim _{n\to \infty}\left|\frac {a_{n+1}} {a_{n}}\right|=\frac 1 e <1$.So absolutely convergent. $\endgroup$ – MatheMagic Dec 26 '16 at 13:31
  • $\begingroup$ @SubratKumarJena Yup. $\endgroup$ – Simply Beautiful Art Dec 26 '16 at 13:32
  • 1
    $\begingroup$ Clear and concise. (+1). $\endgroup$ – Mark Viola Dec 26 '16 at 16:29


(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?

  • 1
    $\begingroup$ Clear and concise. (+1) $\endgroup$ – Mark Viola Dec 26 '16 at 16:30

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