2
$\begingroup$

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

$$\sum\frac{1}{2+3^{-k}}$$

I did the following

$$\sum\frac{1}{2+3^{-k}}<\sum\frac{1}{3^{-k}}$$

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.

$\endgroup$
  • 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
7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.