How would one find if the following series is divergent or convergent.


I did the following


But I am not sure what test I should use the only ones I know are limit comparison and basic comparison test.

But what should I do.

  • 3
    $\begingroup$ The terms do not go to $0$. $\endgroup$ – André Nicolas Oct 25 '13 at 20:36
  • 1
    $\begingroup$ How about $\frac1{2+3^{-k}} \ge \frac13$? $\endgroup$ – MJD Oct 25 '13 at 20:36
  • $\begingroup$ Hint: $\forall k\gt 0,\frac 13\le \frac 1{2+3^{-k}}\le \frac 12$... $\endgroup$ – abiessu Oct 25 '13 at 20:36

Hint: $$\lim_{k \to +\infty} \frac{1}{2+3^{-k}} = \frac12.$$

  • $\begingroup$ It goes to 1/2 but how? $\endgroup$ – Fernando Martinez Oct 25 '13 at 20:38
  • $\begingroup$ $3^{-k}\to 0\text{ as }k\to \infty$ $\endgroup$ – abiessu Oct 25 '13 at 20:39
  • $\begingroup$ oh yes I forget because it is $\frac{3}{1^k}$ k approach infinity $\endgroup$ – Fernando Martinez Oct 25 '13 at 20:41
  • 2
    $\begingroup$ So it is divergent because when you take the limit as k approach infinity it is not zero. $\endgroup$ – Fernando Martinez Oct 25 '13 at 20:42
  • $\begingroup$ Yes @FernandoMartinez: the limit of the series' sequence is not zero and thus the series cannot converge... $\endgroup$ – DonAntonio Oct 25 '13 at 21:25

It's of course divergent because the item of a convergent series must approach $0$ as $n\to \infty$.


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