Math 2260: Calculus II For Science And Engineering
Fall 2013
Concept Review for Exam 3: Sequences and Series
This document briefly outlines the topics which can appear on this exam. Along with
these topics, we ask questions for you to evaluate how well you understand these topics. It’s
a great idea to discuss some of these questions with your teacher or with your friends; they
will also help you pick practice problems to review.
This will have some formulas on it, but do not expect this to be a complete list of formulas
you may need.
Don’t forget the handouts available on the class site and eLC for extra info!
• Sequences
– Recursive sequences (not on test)
– Factorials
∗ 0! = 1, 1! = 1, 2! = 1 · 2, 3! = 1 · 2 · 3... In general, n! = (1)(2) · · · (n).
∗ How do we simplify ratios involving factorials?
– Recap of some forms of limits
∗ L’Hôpital’s Rule: If a limit has 0/0 or ∞/∞ form, then
f (n) L0 H
f 0 (n)
= lim 0
n→∞ g(n)
n→∞ g (n)
lim
∗ If you have ∞ · 0 form (like n ln(1 + 1/n)), make a fraction by flipping something into the denominator. (For instance, ln(1 + 1/n)/(1/n))
∗ If an has an indeterminate power like ∞0 or 1∞ , find limit of ln(an ) first.
Lastly,
ln(an ) → L leads to an → eL
∗ When you take a logarithm above, what form does your limit become? Why?
– The dominant-term approach: To find the limit of a ratio of powers, factor out
the dominant terms to try and get cancellation and expressions going to 0.
∗ We generally
skip showing all the steps. For instance,
to get the limit of
√
√
3
2
2
2
2
2
(n + 5) n − 1/(2n − n + 1), we factor n and n out of the numerator
as well as 2n3 out of the denominator:
q
√ q
√
5
1
5
2
3
2
n (1 + n2 ) n 1 − n2
n (1 + n2 ) 1 − n12
(1 + 0) 1 − 0
1
=
→
=
1
1
1
1
3
3
2(1 − 0 + 0)
2
2n (1 − 2n2 + 2n3 )
2n (1 − 2n2 + 2n3 )
√
though you could just circle the n2 and n2 up top and the 2n3 on the
bottom!
– Sandwich Theorem (not on test)
• Basics of Series
1
– Main definitions
∗ In the series
P∞
n=0
an , the partial sums are
s0 = a0 , s1 = a0 + a1 , s2 = a0 + a1 + a2 , . . .
and in general, sn = a0 + a1 + . . . + an is the sum up to the nth term.
∗ How are these formulas modified when the bottom index is not 0?
∗ The value of the series is limn→∞ sn .
∗ We DO NOT have a formula for the sums sn (except geometric).
– Geometric series:
∗
∞
X
rn =
n=0
1
when |r| < 1
1−r
∗ The ratio r can use x or some other variable (that isn’t n). In this case, solve
|r| < 1 to find the locations of (absolute) convergence.
∗ NOTE: Sometimes solving |r| = 1 instead leads to easier algebra. Why?
∗ Why does a geometric series diverge at its two endpoints?
– How can we adjust the sum of a series when it starts from a different index? Why
does that mean the start index is irrelevant for convergence?
– nth-term test (see series handout)
P
– p-series:
1/(np ) converges when p > 1
R∞
∗ This is very related to the improper integral 1 1/(xp ) dx.
∗ This and the geometric series are the main tools used in things like LCT and
the Ratio Test.
• Limit Comparison (LCT) and Comparison Tests (CT) (see series handout)
– Summary of dominant terms:
ln(n) < np < rn < n! < nn
(where < means “is dominated by” and p > 0, r > 1 here).
– How does the LCT help us simplify summations?
– What purpose does showing an ∼ bn (an /bn → 1) serve?
– When working out an /bn , what can we do to simplify the calculations?
– What scenarios call for CT rather than LCT?
– CT, unlike LCT, consists of two tests that can only do one conclusion each. How
do we know which one we expect to use?
– Why might you need to chain two tests together? (This will be rare for us...)
• Ratio and Root Tests (see series handout)
– Why do we use absolute values in the limits for |r|?
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– When |r| = 1, you get no conclusion from Ratio and Root... what do you try
then?
– Why don’t these two tests help with polynomials and p-series?
– What scenarios make the Root Test easier to use than the Ratio Test? Which
scenarios make the Ratio Test easier?
• Absolute and Conditional Convergence
P
P
– Absolute convergence means
|an | converges (which implies
an converges).
– Checking absolute convergence works just like before with all our series tests!
– The nth-term test still works even with absolute values.
P
P
– Conditional convergence means
|an | diverges but
an converges anyway.
P
– Alternating Series Test: If
an is alternating and |an | → 0, then the series
converges.
∗ Usually, some power of −1 is found in an alternating series like (−1)n or
(−1)n+1 .
∗ When testing an alternating series, FIRST check for absolute convergence.
After that, we consider nth-term test and AST.
• Power Series
P∞
n
– Form:
n=0 an (x − a) where a is the center and each an does not depend on x.
– We first test these using either Root or Ratio Test. (Why?)
– The series converges on an interval centered at a called the interval of convergence
whose radius R is the radius of convergence. There are three possibilities:
∗ R = 0, so the series only converges at x = a.
∗ R = ∞, so the series is always AC.
∗ 0 < R < ∞. The ends of the interval are x = a±R (found by solving |r| = 1).
Between the ends you have AC. Outside the ends, there’s divergence. At the
ends, nobody knows!
∗ When 0 < R < ∞, you plug in the two ends and test separately. These ends
should get exponentials to cancel, and they should be very similar except that
one alternates.
– A lot of similar tactics work with a geometric series that has a ratio involving x.
By finding ends, you get an inside which is AC and an outside (and endpoints)
which diverges. Furthermore, you can find the sum!
• Taylor and Maclaurin Polynomials
– Form of Taylor polynomial: The nth-order polynomial of f (x) with center a is
Pn (x) =
n
X
f (i) (a)
i=0
i!
(x − a)i
In many cases, these coefficients are found by repeatedly computing derivatives
of f .
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– A Maclaurin polynomial is the special case when a = 0, so (x − a)i becomes xi .
– Pn (x) uses terms UP TO the nth power of x.
– The Taylor series (or Maclaurin series) is when you let n → ∞ and get a power
series!
∗ We only compute the full series for certain special functions with easy patterns
for their derivatives.
∗ (See the handout with famous Maclaurin series.)
∗ We can reuse these series to compute some Pn ’s much faster than by repeated
derivatives.
∗ Series of ln(1 + x) and arctan(x) are not on test.
– Remainder Estimation Theorem: This shows how good of an approximation Pn (x)
is.
∗ Suppose x comes from some interval around the center a. Then the Taylor
error |Rn (x)| is bounded as follows:
|Rn (x)| ≤
M |x − a|n+1
(n + 1)!
when M is the biggest value of |f (n+1) (x)| on the interval.
∗ You do not have to memorize this formula.
∗ Usually, you find some derivatives in getting Pn (x) anyway. Then you compute (n + 1)st derivative, take its absolute value, and figure out which point
in the interval is going to give the biggest value. (Often, it comes down to
whether |f (n+1) | is increasing or decreasing.)
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