If the general term of the sequence is written as $b_n\={\left(-1\right)}^n \frac{2^{-n}}{n}$, the sequence will be recognized as an alternating sequence because $a_n\=\left|b_n\right|\=2^{-n}\/n\>0$ for all $n\ge 1$. If the sequence of absolute values converges, then the alternating sequence is both absolutely and conditionally convergent. Hence, apply L'Hôpital's rule to obtain

$\underset{n\to \infty }{lim}\frac{2^{-n}}{n}\=\underset{n\to \infty }{lim}\frac{-2^{-n}\ln \left(2\right)}{1}\=-\ln \left(2\right) \underset{n\to \infty }{lim}\frac{1}{2^n}\=0$ 

The given alternating series converges absolutely. This example is included to remind the reader that not every convergent alternating sequence converges conditionally. An alternating series can converge absolutely, as this example shows.

Table 8.1.3(a) contains the  task template that, given the general term of a sequence, calculates and graphs its first few members.

Place the cursor somewhere in the cell containing the phrase "General term"and press the Tab key often enough for the cursor to move to, and select the default general term. With this expression auto-selected, simply overwrite with the desired general term, most easily obtained by a copy/paste operation. Then, adjust any of the inputs as needed, and simply press the Enter key to execute each command in the template.

The astute reader will by now have realized that the Maple calculations above simply show that the sequence $\left\{b_n\right\}$ converges. They do not show that the sequence converges absolutely. The following calculation establishes the absolute convergence.