# Finding if $\sum\frac{1}{2+3^{-k}}$ divergent or convergent

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.

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

## 2 Answers

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

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

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