General Contents
Detailed Contents
Index
Programmed tutorial: Alternating Series: Convergence or Divergence: Example 1
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If you want to see all of the following steps at once, click the "All Steps" button. Otherwise, use the "Next" button.
Test whether the alternating series
converges.
General Contents
Detailed Contents
Index
In order to visualize this series, plot its terms on paper.
Then check your plot by clicking “Next”.
alt="Java applet graph of the terms of the alternating series." Your browser is completely ignoring the <APPLET> tag!
To get further
insight,
plot the sequence of partial sums on paper.
Then check by clicking “Next”.
alt="Java applet graph of the partial sums of the alternating series." Your browser is completely ignoring the <APPLET> tag!
Does it appear that the series is converging?
Yes, the partial sums appear to be approaching approximately -0.06.
How can we show that this series is convergent?
Since it is an alternating series, we can try to show that
for
n
≥ 1,
and that
.
First, let’s use the general Comparison Test.
Set it up.
Do both the numerator and the denominator each approach ∞?
Yes.
How do we
proceed
?
We need to apply
L’Hopital’s
Rule.
Describe this rule.
We differentiate the numerator and the denominator separately, take the ratio, and evaluate the limit of the resulting expression. (We do not use the Quotient Rule).
We need a continuous function of
x
for L'Hopital's Rule.
What function should we use?
We can
use
, and take
Apply
L’Hopital’s
Rule.
Differentiating the numerator and denominator separately, we get
Can we find the limit now?
Yes, only the denominator approaches ∞ now.
What can we conclude is the limit L?
The limit is
L
= 0.
Is this condition also satisfied for
a
n
?
Yes.
Next, use the definition of
a
n
in the
inequality
.
All the
a
n
are positive, so we
get
Simplify by multiplying both sides by 3
n+1
.
We
get
.
Solve the inequality.
We can subtract
n
/3 from both sides:
, or
, which is true for
n
≥ 1
What is our conclusion?
Both conditions are satisfied for this alternating series to converge.
Now let’s find the remainder after summing N terms.
State the remainder informally.
The remainder involved in approximating
S
by
S
N
is smaller than the first neglected term.
Write
this in terms
of an inequality.
To get the remainder when using 5 terms as an approximation, rewrite this for N = 5.
Set up the determination of
S
5.
Determine
S
5.
Determine
a
6
.
Use these results to get bounds on
S
.
, or
Combine these into a three-part inequality.
We get
Does
this agree
with our earlier diagram?
Yes, the partial sums appeared to approach -0.06 on the diagram.
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