Chapter 8: Infinite Sequences and Series
Section 8.1: Sequences
If bn=−2−nn, find the limit of the sequence bnn=1∞.
If the general term of the sequence is written as bn=−1n 2−nn, the sequence will be recognized as an alternating sequence because an=bn=2−n/n>0 for all n≥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
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.
Calculus palette: Limit operator
Context Panel: Evaluate and Display Inline
limn→∞−2−nn = 0
Table 8.1.3(a) contains the
task template that, given the general term of a sequence, calculates and graphs its first few members.
First index value
Last index value
plotseq,expr,=.., style=point, symbol=solidcircle, color=red
Table 8.1.3(a) The Sequences task template
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 bn converges. They do not show that the sequence converges absolutely. The following calculation establishes the absolute convergence.
limn→∞2−nn = 0
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