General Contents
Detailed Contents
Index
Programmed tutorial: Telescoping Series: 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.
Find the sum:
General Contents
Detailed Contents
Index
Plot this series on paper.
Then click “Next” to check your graph.
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Does the series appear to be converging?
Yes
Can we factor the expression for the nth term?
Yes
Rewrite the series with factors in the denominator.
What technique does this suggest?
A partial fraction expansion and an attempt at “telescoping” the series.
Set up the expansion for the nth term.
How do we determine A and B?
We add the two fractions, simplify, and equate coefficients of each power of
n
.
What will the Least Common Denominator be?
The LCD is
Are we “going in circles” here?
In a way, but we will be able to determine A and B.
Rewrite the fractions, using the LCD.
We get
Do the multiplications.
Collect like terms in n.
Factor in those groups.
How do we solve for A and B?
We set this equal to the original factorization.
Do that.
What do we do now?
We equate the coefficients of like terms on each side.
Equate the coefficients of n.
How?
There is no coefficient of n on the right.
We take that coefficient of n on the right as 0.
We get
, or
, or
Equate the constant terms from each side.
, or
Substitute
A = -B
from the previous result.
, or
, or
Solve for
A
in
A = -B
Combine these results to rewrite
as the sum of two terms.
Plot these two terms on paper and click on “Next” to confirm your graph.
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Notice that some blue points are symmetric with some of the red points, indicating that there may be “telescoping”.
What property of the sum of two terms can we apply?
We can replace the sum by two sums and factor out the ¼.
Rewrite S using these properties.
Write out the first few terms of each series to see if there is any “telescoping”.
What constitutes the “telescoping” here?
The fact that from the term
on to ∞, there is a negative term in the first sum that eliminates the corresponding term in the second sum.
Here is a diagram with green lines that connect the points which eliminate.
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What do we do with the remaining terms
We add them to get the sum, S.
To add those fractions, what is the Least Common Denominator?
The LCD is 12.
Set up the addition of those fractions.
Calculate the resulting value of S.
Summarize the method demonstrated here.
To sum a series with a factorable denominator, try a partial fraction expansion of the n
th
term.
Then look for elimination of corresponding terms (telescoping).
Add the remaining terms to get the sum of the original series.
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