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Sequences and Series
Collège André-Chavanne
Genève
« [email protected]
2012
2
1
INFINITE SUMS
1
Infinite sums
The question is: can an unending sum give a result? Does
∞
X
1
2k
k=0
have a meaning?
...
1
32
1
64
1
8
1
16
1
2
1
4
Exercise 1
Is it correct to write
1
1 1
1 1 1
x= + + +
··· = + ·
2 4 8 16
2 2
hence x =
1
2
1 1
1
+ +
...
4 8 16
=
1 1
+ ·x
2 2
+ 12 x and therefore x = 1?
Exercise 2
Using the same method as above, calculate:
x = 1 + 2 + 4 + 8 + 16 + . . .
The question is: what went wrong? In order to answer this sort of question, we will first
study another type of unending process.
3
2
2
SEQUENCES
Sequences
Definition 1
A sequence is a function
u : {k, k + 1, . . . } ⊆ N −→ R,
with k ∈ N.
We also use the notation:
(un )n≥k
for the sequence above, with un = u(n), for n ≥ 0. We also write (un ) if the set of indices is
obvious or irrelevant. The numbers un are called the terms of the sequence.
An observation level of a sequence is the level containing the parameters used in its definition,
in particular it contains the integer k.
A finite sequence can be given by enumeration:
1, 2, 3, 4, 5
If the rule for obtaining the elements is obvious, dots will be used:
1, 2, 3, . . . , 20, 21, 22
If the sequence is infinite, the enumeration is impossible, but if the rule is obvious, dots will
be used to indicate the never ending succession:
1, 2, 3, 4, 5, . . .
Sets are noted with braces {...} If the general term of the sequence can be represented by
an the notation {an }n is be used.
The first and last elements of a sequence will be indicated by superscripts and subscripts:
{an }20
n=1
is the notation for the sequence a1 , a2 . . . a19 , a20
Exercise 3
Find {an }n for the sequence of all natural numbers
For non ending sequences, the symbol {n}∞
1 can be used to indicate that its number of terms
1
exceeds all finite terms. ( )
Exercise 4
Write the first terms of the following sequence:
∞
1
n 1
Exercise 5
Write down the first five terms of the sequences specified by their nth terms (in each case,
n ∈ N)
1
The ∞ sign expresses the idea of an unending process. This symbol does not represent a number, not even an
ultralarge number. It means “never ending”
4
2
SEQUENCES
(1) un = 4n
(4) bn = 2n2 − 1
(2) tn = 2n−1
(5) rn = (−1)n
(3) an = 3n − 2
(6) en = (−1)n
n2
n+1
Definition 2 (Explicit Relation)
An explicit relation expresses the kth term as a function of k.
Definition 3 (Recurrence Relation)
A recurrence relation expresses the kth element of a sequence in terms of one or more of its
predecessors.
In order to know where the sequence begins, it is necessary to state the value of the first (or
first few) terms of the sequence. a
a
Recall the conditions for an induction proof.
Exercise 6
Write the first terms of the sequences:
uk−1
(1) u1 = 1 uk =
k+1
(2) u1 = 1 un = 2 · un−1
(3) u1 = 2 ui = 3 + 2ui−1
If possible, rewrite them in explicit form. Why do you need an induction proof?
A Graph of a sequence helps to see how it behaves. Joining the plotted dots by a dotted
line (because the domain is defined on natural numbers, the plot will be discrete points)
The sequence {1, −1/2, 1/3, −1/4, 1/5, −1/6, 1/7, −1/8, } is plotted below:
Exercise 7
Give the first terms of the following sequences:
un−1
(1) u1 = 5 un = 1 +
10
1
(2) u1 = 0 un =
5 − un−1
If possible, rewrite them in explicit form.
5
2
SEQUENCES
Exercise 8
Write the first terms of the following sequences:
(1) u1 = 0 u2 = 1 ur = 2ur−1 − ur−2
(2) u1 = 1 u2 = 3 uk = 3uk−1 − 2uk−2
Example: Let a and d be two real numbers and let k be a positive integer. We define an
arithmetic progression (with common difference d) as follows:
uk = a and un+1 = un + d for n ≥ k.
It is immediate that un = a + (n − k) · d, for all n > k. The context of this sequence is given by
a, d, k.
Example: In a similar way, given a, r ∈ R and k ∈ N, we define a geometric progression (with
common ratio r) by
uk = a and un+1 = un · r for all n > k.
Then un = a · rn−k for all n < k. A context of this sequence is given by a, r, k.
Definition 4 (Observation level of a sequence)
The observation level of a sequence {an }n is given by its parameters, not by n which is the
variable.
Definition 5 (Convergence)
Let (un )n≥k be a sequence. We say that (un )n≥k converges if lim un exists i.e., if there is an
n→∞
L ∈ R such that
lim un = L.
n→∞
The number L is the limit of the sequence.
or
(un )n≥k converges if there is an observable L ∈ R such that for all ultralarge N
uN ' L.
Definition 6 (Non convergence)
A non-convergent sequence may be
• divergent (the terms eventually get ultralarge),
• periodic or oscillating. (values repeat after a cycle of constant length.)
!
A sequence does not necessarily “reach” its limit.
6
2
SEQUENCES
Exercise 9
What are the conditions for an arithmetic sequence to converge?
What are the conditions for a geometric sequence to converge?
Exercise 10
(1) Find the limit of
1
(2) Find the limit of
n
n n
n
n+1
o
n
Exercise 11
Describe the behaviour of the following sequences:
(1) {(−1)n }n
(2) u1 = 1 un+1 = 1 −
(3)
1
1 + un
cos n π3 n
(4) {random numbers between -1 and 1}
(5) un = n(n + 1)(n + 2)
(6) {n}n
(7) u1 = 1 u2 = 1 un =
un−1
un−2
Exercise 12
One of the most famous sequences: The Fibonacci sequence.
2
u1 = 0 u2 = 1 un = un−1 + un−2
(1) Write the first terms (at least ten) of the sequence and describe the behaviour of the
sequence.
(2) Make a new sequence vk with the following rule (with un the sequence just calculated),
sketch the first terms and describe the behaviour.
u2
u3
u4
un+1
v1 =
v2 =
v3 =
vn =
u1
u2
u3
un
(3) Use the same rule as in the beginning, but start with any two numbers for u1 and u2 (even
with u2 > u1 ) and calculate the second sequence vn made from these terms and sketch
the first terms. Describe the behaviour.
2
Fibonacci was an Italian mathematician in the XIIþ century; it was he who introduced Arab-Indian numerals
into Europe.
7
2
(4) Write the first terms of the sequence: w1 = 1
the sequence.
wn = 1 +
1
wn−1
SEQUENCES
Describe the behaviour of
Definition 7
The sequence (un )n≥k is:
(1) increasing if un ≤ um for all m ≥ n ≥ k,
(2) decreasing if un ≤ um for all m ≥ n ≥ k,
(3) monotone if (un )n≥k is either increasing or decreasing.
(4) bounded above if there is an M ∈ R such that un ≤ M for all n ≥ k (the number M is
an upper bound),
(5) bounded below if there is an M ∈ R such that un ≥ M for all n ≥ k (the number M is a
lower bound),
(6) bounded if the sequence is either bounded above and bounded below.
Let (un )n≥k be a sequence. If it is bounded above then by the context principle there is an
M appearing at the observation level which is also an upper bound. Conversely, if there is an
M appearing at the observation level such that
un ≤ M, for all n appearing at the observation level,
then by the context principle this statement is true for all integers (including ultralarge integers).
The same remark holds for lower bounds.
Definition 8 (Least Upper Bound)
A least upper bound M to a nonempty set A of real numbers is a value such that x > M ⇒ x ∈
/A
and for any N < M there is an x ∈ A such that x > N .
A similar definition holds for greatest lower bound.
Exercise 13
Theorem 1
A set of real numbers bounded above has a least upper bound (l.u.b). A set of real number
bounded below has a greatest lower bound (g.l.b)
This theorem is not true if one replaces "real" by "rational. Consider
{p/q ∈ Q | (p/q)2 < 2}
Why does this not have a a least upper bound?
• Why is it possible to state that there is an observable a ∈ A and there is an observable
upper bound B?
• Let N be ultralarge. Divide the interval [a, B] in N even parts.
8
3
SERIES
• Try to finish the proof of the theorem.
You will need to use closure in two versions: the usual one
"If there is an x satisfying a property, then there is an observable x satisfying that property."
and its contrapositive
"If all observable x satisfy a property, then all x satisfy that property."
Exercise 14
prove the following theorem:
Theorem 2 (Monotone Convergence)
Any increasing sequence which is bounded above is convergent and has a limit. Similarly, any
decreasing sequence which bounded below is convergent and has a limit.
Definition 9
Let (un )n≥k be a sequence. We say that (un )n≥k is a Cauchy sequence if
uN 0 ' uN ,
for all positive ultralarge integers N, N 0 .
A context is given by the sequence. By the context principle, a sequence is a Cauchy
sequence if and only if this condition is met for any level finer than the observation level.
Exercise 15
Prove the following theorem:
Theorem 3
Let (un )n≥k be a sequence. Then (un ) converges if and only if (un )n≥ is a Cauchy sequence.
Exercise 16
Back to Fibonacci. Use theorem 3 to prove that the sequence of ratios converges. Show that
the recurrence relation has a fixed point. Then show that this fixed point is the limit.
3
Series
Let (un )n≥k be a sequence. It is possible to define another sequence by considering the partial
sums sk = uk and sn+1 = sn + un+1 , for n ≥ k. In other words, for a positive integer N we
have
N
X
sN = uk + uk+1 + · · · + uN =
un .
n=k
9
3
SERIES
Definition 10 (Partial Sum)
A partial sum is the sum up to a given index number. It is indicated by
S1 =
1
X
S2 =
2
X
i=1
Sn =
i=1
n
X
i=1
Definition 11 (Infinite Series)
An infinite series is the limit of its partial sums.
An infinite series has a sum iff it converges.
Let (un )n≥k be a sequence. A series is the sequence
N
X
!
un
n=k
N ≥k
of the partial sums. We will denote this series by
∞
X
un .
n=k
This definition is equivalent to:
Definition 12 (Convergence of a Series)
A series converges iff there is an observable L such that for any ultralarge N
N
X
ui ' L
i=k
Partial sums represent successive approximations of the total sum
uk + uk+1 + uk+2 + . . .
which is not necessarily a real number in the sense that it is not guaranteed that the partial
sums converge.
Definition
13
P
P
Let n≥k un be a series. We say that n≥k un converges to L if the sequence of patial sums
converge to L.
If the series converges, then the total sum is equal to the limit of the sequence of partial
sums. As before, if the limit exists, it appears at the observation level of the series (or of the
sequence).
Exercise 17
Here is a well known series for which it may be possible to guess the limit. However, the
question is how to prove it.
Calculate
∞
X
1
2i
i=0
10
3
SERIES
Exercise 18
Same question (also graphically) for
∞
X
1
4k
k=1
Exercise 19
Same question for
∞
X
1
nk
k=1
P
Example: Consider the arithmetic series n≥1 un , with u1 = a and un = a + (n − 1) · d.
P
A context is given by u. To establish the value of N
n=1 un first note that
1
N · (N − 1)
1 + 2 + · · · + N − 1 = (1 + (N − 1) + (2 + (N − 2)) + · · · + (N − 1) + 1)) =
2
2
thus
N
X
a+(n−1)·d = N ·a+d
n=1
N
X
n−1 = N ·a+d·
n=1
N
N
N · (N − 1)
= (2a+(N −1)d) = (u1 +uN ).
2
2
2
If N is ultralarge, then
N
X
un =
n=1
N
(2a + (N − 1)d)
2
is also ultralarge.
Hence an arithmetic series cannot converge unless a = d = 0.
P
Example: Consider the geometric series n≥1 un , with u1 = a and un = a · rn−1 , with a, r ∈ R
P
(a 6= 0). Let sN = N
n≥1 un . Note that:
sN + arN +1 = a + ar + ar2 + · · · + arN + arN +1 = a + r(a + ar + · · · + arN ) = a + r · sN
therefore sN · (1 − r) = a · (1 − rN +1 ) and
sN = a ·
1 − rN +1
,
1−r
if r 6= 1.
Note that if r = 1 then sN = a · N so with a common ratio equal to 1, if the initial term is not
zero, the series diverges.
Ut is simple to check that
(
X
diverges if |r| ≥ 1
a · rn
1
if |r| < 1.
converges to a · 1−r
n≥1
The next exercise shows that the initial terms do not influence the convergence of a series.
11
4
CONVERGENCE CRITERIA
Exercise
P20
P
Let n≥k P
un be a series and m an integer such that m ≥ k. Show that n≥k un converges
if and only if n≥m un converges.
Exercise 21
For a geometric series with 2 as first term and r = 3/4. Write the first terms. Calculate the
limit.
Exercise 22
Let
∞
X
3 · 10−k
k=0
Does this series converge and if so, what is its limit?
Exercise 23
Calculate (if the value exists) of
∞
X
0.999j
j=0
4
Convergence Criteria
Exercise 24
Use theorem 3 to prove the following.
Theorem
P 4
P
Let n≥k un be a series. Then n≥k un converges if and only if
0
for any ultralarge numbers N < N
0
N
X
un ' 0
n=N
Theorem 5 (Comparison test)
Let (un )n≥k and (vn )n≥k be two sequences with non-negative terms such that
for each n ≥ k.
P
P
If the series n≥k un converges then the series n≥k vn converges also.
un ≥ vn ,
The contrapositive of the previous theorem can be used to prove the divergence of a series.
Theorem
6
P
If n≥k un converges then limn→∞ un = 0.
12
4
CONVERGENCE CRITERIA
Example: The converse of this theorem is false: consider the harmonic series
X1
.
n
n≥1
1
= 0. We will show now that this series diverges.
We have lim
n→∞ n
We observe that
1
1 1
1
1 1
1 1
1
≥1+ +
=1+ + =1+2·
s4 = 1 + +
+
+
2
3 4
2
4 4
2 2
2
1
1 1 1 1
1
1
≥ s4 + 4 · = s4 + ≥ 1 + 3 · .
s8 = s4 + 1 +
+ + +
5 6 7 8
8
2
2
By induction, we see that
1
S2N ≥ 1 + N · .
2
But this implies that the series diverges because if N is ultralarge then 2N is ultralarge and
S2N ≥ 1 + N2 is ultralarge hence not at the observation level.
Theorem 7 (Integral Test)
RN
Let f : [k, ∞[→ P
R be a continuous decreasing and positive function. Let F (N ) = k f (x) · dx.
Then the series n≥k f (n) converges if and only if limN →∞ F (N ) exists.
Example: Consider the series
decreasing function. Then
1
n≥1 n2 .
Let f :]0, ∞[ by x 7→
P
F (N ) =
1
Then limN →∞ F (N ) = 1 so the series
It is a positive continuous
1
1 N
1
dx = − = 1 −
2
x
x 1
N
N
Z
1
.
x2
1
n≥1 n2
P
converges.
Exercise 25
The Riemann series is
X 1
,
np
with p ∈ R.
n≥1
Show that the Riemann series converges if and only if p > 1.
The two following criteria use comparisons with some geometric series.
Theorem
P 8 (Ratio Test)
Let n≥k un be a series with strictly positive terms.
If
un+1
lim
=L
n→∞ un
then
(1) if L > 1 the series diverges,
(2) if L < 1 the series converges.
13
5
SERIES FOR THE EXPONENTIAL
The ratio test is inconclusive in the case L = 1: we have seen that
P
1
n≥1 n2 converges.
1
n≥1 n
P
diverges but
Definition
P14
A series k≥n un (or (un )n≥k ) is an alternating series, if un · un+1 < 0 for each n ≥ k.
Theorem 9
P
Let (un )n≥k be an alternating series decreasing in absolute value. If lim un = 0 then n≥k un
n→∞
converges.
Example: This shows that the harmonic alternating series defined by
X
1
(−1)n
n
n≥1
converges.
Exercise 26
P
k
Show that if one considers the series ∞
k=1 (−1) then by rearranging the order of the terms,
the sum can be made to be equal to any positive or negative number.
! This is a crucial point. A never ending series can yield strange things! Because it never
ends. This is why it is important to work on the partial sums. (More difficult theorems state
under what conditions can the terms of a series be rearranged without changing the result.)
5
Series for the exponential
Exercise 27
Using any of the convergence criteria, prove that
∞
X
xk
k=0
k!
converges for any value of x.
Since it ocnverges and depends on x, we define
f : x 7→
∞
X
xk
k=0
k!
Calculate f 0 (x).
Exercise 28
Same idea for the product; show that this "infinite product" converges, then differentiate it.
x n
g : x 7→ lim 1 +
n→∞
n
14
6
THE LOGARITHM FUNCTION
Exercise 29
Show that the two functions given above (f and g) are the same. Hint: differential equations
Exercise 30
Show that
∞
X
12 13
1 n
1
=2+
1+
+
+ ··· =
n→∞
n
2!
3!
k!
lim
k=0
Definition 15
x
e =
∞
X
xk
k=0
k!
= lim
n→∞
1+
x n
n
Exercise 31
Compute the first partial sums for ex with x = 1 and other values of x and compare with the
x
e value of your calculator.
Exercise 32
From complex numbers, recall that eix = cos x + i sin x
(1) Write the beginning of the series for eix
(2) Write the series for cos(x) and sin(x) (Think about the real part and imaginary part
separately).
(3) Prove that these series converge.
(4) Calculate cos(1) using this series.
(5) Calculate tan(1).
6
The Logarithm Function
ln0 (x) =
1
x
Can we find a series which when differentiated yields x1 ? This would be the series for ln(x).
It proves to be very difficult, so we will go to the series for ln(1 + x) where, through the chain
1
1
rule, ln0 (1+x) =
, so the question becomes: what series yields
when differentiated?
1+x
1+x
or: what series, when multiplied by (1 + x) yields 1?
15
6
THE LOGARITHM FUNCTION
Exercise 33
Suppose the series to be of the form a0 + a1 x + a2 x2 + · · · + an xn + . . .
Multiply it by 1 + x and collect similar terms.
This must be equal to 1 (hypothesis) so all powers of x must disappear.
Write the series for ln(2).
1
The figure shows the graph of f : x 7→ 1+x
and some of the partial sums, noted by Sn When
the value of n is high, the series converges on a longer interval of [0; 1] but the faster it gets
away afterwards.
S3
S5
S1
1/(1 + x)
S4
S2
!
1
The assumption that 1+y
can be written as the sum of powers of x is not true for x > 1.
1
This is why the different partial sums approach the function x 7→ 1+x
only on a short interval.
Exercise 34
With the series above, calculate the series which when differentiated, yields
find the series for ln(1 + x). Prove that it converges if x < 1.
As with the series for
1
1+x
1
1+x
and thus
the series for ln(1 + x) converges only for x ∈ [0, 1]
Exercise 35
Use the product definition and the series for ln(1 + x) to show that ln(ex ) = x
Remember the fundamental property of the logarithm, that the logarithm of a product is the
sum of the logarithms.
16
7
7
TAYLOR SERIES
Taylor Series
P
The idea of this part is to represent a function by a series n≥k an · (x − c)n such that the
series converges to f (x) for some values of x around a point c. This is called the Taylor series
for f at c.
We first define the nth derivative of f by induction on n.
Definition 16
Let f be a function. We say that f is differentiable once at x if f 0 (x) exists. We write
f (1) (x) = f 0 (x). By induction, for a positive integer n, we say that f is differentiable n + 1
times at x if the function f (n) is differentiable at x. We write f (n+1) (x) = (f (n) )0 (x).
Theorem 10
Let N be a positive integer and let c ∈ R. Let f be a function differentiable N + 1 times on an
open interval containing c and let x be in this interval. Then
f (x) =
N
X
(x − c)n
n!
n=0
·f
(n)
Z
(c) +
c
x
(x − t)N
· f (N +1) (t) · dt.
N!
We give a first application:
Theorem 11
The number e satisfies
e = lim
n→∞
1
1
1
1 + + + ··· +
1! 2!
n!
.
We have already shown that the alternating harmonic series converges. Now we can show
more.
Exercise 36
Prove that the alternating harmonic series
X (−1)n+1
n≥1
ln(2) = lim
n→∞
1−
n
converges to ln(2) i.e.,
(−1)n+1
1 1 1
+ − + ··· +
2 3 4
n
.
Exercise 37
Prove:
Theorem 12
Let N be a positive integer and c ∈ R. Let f be a function N + 1 times differentiable on an
open interval containing c and let x be in this interval. Then
f (x) =
N
X
(x − c)n
n=0
n!
· f (n) (c) +
(x − c)N +1 (N +1)
·f
(ξ),
(N + 1)!
17
with ξ between c and x.
7
TAYLOR SERIES
This observation gives us a convergence criterion for these series:
Theorem 13
Let f be a function infinitely many times differentiable on an open interval containing c and let
x be in that interval. Suppose that there exists an M such that for each positive integer n the
function f (n) is bounded by M on [x; c] (or [c; x] if c < x). Then the series
X (x − c)n
n≥0
n!
· f (n) (c) converges to f (x).
Theorem 14
X xn
For each real number x the series
converges to exp(x) i.e.,
n!
n≥0
exp(x) = 1 + x +
xN
x2 x3
+
+ ··· +
+ ....
2!
3!
N!
Exercise 38
For each of the following, calculate the first terms of the Taylor series. Use induction to
obtain the general term. Prove that it converges. Use c = 0 for all three.
(1) cos(x)
(2) sin(x)
(3) arctan(x)
Using a calculator, compute f (1). What is this value?
Example: A Taylor series for f may converge everywhere without converging to the function f .
Consider f given by
( 1
e− x2 ,
if x 6= 0
f (x) =
0
otherwise.
One can show that
f (n) (0) = 0,
for each positive integer n.
The power series n≥0 0 · xn converges to the function which is everywhere 0 and not to f .
This is not a contradiction to Theorem 13, as for each x 6= 0 and each M there exist n and ξ
between 0 and x such that |f (n) (ξ)| > M , so the assumptions of the theorem are not satisfied.
P
Exercise 39
√
Calculate the Taylor series for x. You must first find a good value for c, which might mean
trying several values.
Does it converge for all values of x? (Try using to compute square roots of 0,1,4...)
If it does not converge for, say, 10, is it possible to use another value for c?
18
7
TAYLOR SERIES
Practice exercise 1 Answer page 21
For the following, find the partial sums, determine whether the series converges and find the
sum when it exists.
n
1 1
1
(1) 1 + + + · · · +
+ ...
3 9
3
n
3
9
3
(2) 1 + +
+ ··· +
+ ...
4 16
4
1 1
1
1
1
1
1
+
+
+ ··· +
+ ...
(3) 1 −
−
−
−
2
2 6
6 24
n! (n + 1)!
(4)
1
1
1
+
+ ··· +
+ ...
1·2 2·3
n(n + 1)
Hint:
1
n(n+1)
=
1
n
−
1
n+1
1
n2
−
1
(n+1)2
(5) 1 − 2 + 4 − 8 + · · · + (−2)n + . . .
(6)
12
3
5
2n + 1
+ 2 2 + ··· + 2
+ ...
2
·2
2 ·3
n (n + 1)2
Hint:
(7)
1
1
1
1
+
+
+ ··· +
+ ...
1·3 3·5 5·7
(2n − 1) · (2n + 1)
(8)
1 2 3 4
(−1)n−1 · n
− + − + ··· +
3 5 7 9
2n + 1
(9)
1 1
1
1
+ +
+ ··· +
+ ...
4 7 10
3n + 1
2n+1
n2 (n+1)2
=
(10) ln(1) + ln(2) + ln(3) + · · · + ln(n) + . . .
Practice exercise 2 Answer page 21
For the following, the general term of the series is given. Test the corresponding series for
convergence:
(1)
3n − 7
10n + 9
5
2
6n + n − 1
√
n
√
(3)
1+2 n+n
(2)
(4) n · e−n
(5)
5n
3n + 4 n
(6)
nn
(n!)2
(7)
2n · n!
nn
(8)
1
ln(n)
(9)
n2
2n
(10)
19
ln(n)
n
7
TAYLOR SERIES
Practice exercise 3 Answer page 22
Give the Taylor series for the following. State for which values of x they converge.
(1)
1
1−x
(6) e−x
2
(7) e−x
Z x
2
(8)
e−t dt
1
(2)
1+x
(3)
1
1 − 2x
0
(9) ln
(4) ln(1 − x)
(5)
1
1 + x2
1+x
1−x
(10) (1 + x)p for fixed p.
20
7
TAYLOR SERIES
Answers to practice exercice 1, page 19
n 3
1
(1)
1−
. Converges to 23 .
2
3
n 3
(2) 4 1 −
. Converges to 4.
4
(3) 1 −
1
. Converges to 1.
(n + 1)!
(4) 1 −
1
. Converges to 1.
n+1
(5) If n is even: −n/2. If n is odd: n/2 + 1/2. Diverges.
1
. Converges to 1.
(n + 1)2
1
1
1−
. Converges to 12 .
(7)
2
2n + 1
(6) 1 −
1
(8) Diverges because lim an = .
n→∞
2
(9) Diverges because for ultralarge N , S2N 6' S2·2N .
(10) Diverges because for ultralarge N , ln(N ) 6' 0.
Answers to practice exercice 2, page 19
(1) Diverges since for ultralarge N we have
3 − 7/N
3
3N − 7
=
'
6' 0
10N + 9
10 + 9/N
10
5
5 1
<
+n−1
6 n2
√
Z ∞
R∞
√
x
√
(3) Converges. By integral test:
dx. Substitute x = u and 0
2
( x + 1)
0
converges.
an+1
1 1
(4) lim
= 1+
<1
n→∞ an
n e
(2) Converges. By comparison:
(5) Diverges. By ratio test:
6n2
1
du
(u+1)2
5(32 + 4n )
5(32 + 4n )
5
>
= < 1.
n
n
n
n
3·3 +4·4
3·3 +3·4
3
(6) Converges. By ratio test:
(n + 1)n
(nn (1 + 1/n)n
e
=
'
→ 0.
n
n
(n + 1)n
(n + 1)n
n+1
(7) Converges. By ratio test: '
2
e
< 1.
(8) Diverges. By comparison: ln(n) < n hence
21
1
1
> and the harmonic series diverges.
ln(n)
n
7
TAYLOR SERIES
an+1
1
= < 1.
n→∞ an
2
(9) Converges. Ratio test: lim
ln(x)
1 2 ∞
(10) Diverges. Integral test: by setting ln(x) = u we get
dx = ln (x) which
x
2
1
1
diverges. By horizontal shifting, this function is below the series.
Z
∞
Answers to practice exercice 3, page 20
(1) 1 + x + x2 + x3 + x4 + . . . for |x| < 1
(2) 1 − x + x2 − x3 + x4 − . . . for |x| < 1
(3) 1 + 2x + 22 x2 + 23 x3 + 24 x4 + . . . for |x| < 1/2
(4) −x −
x2 x3 x4
−
−
− . . . for |x| < 1
2
3
4
(5) 1 − x2 + x4 − x6 + x8 − . . . for |x| < 1
(6) 1 − x +
x2 x3 x4
−
+
− . . . for all x
2!
3!
4!
(7) 1 − x2 +
(8) x −
x4 x6 x8
−
+
. . . for all x
2!
3!
4!
x3
x5
x7
x9
+
−
+
− . . . for all x
3!
5 · 2! 7 · 3! 9 · 4!
(9) 2x +
2x3 2x5 2x7 2x9
+
+
+
+ . . . for |x| < 1
3
5
7
9
(10) 1 + px +
p(p − 1) 2 p(p − 1)(p − 2) 3
x +
x + {p4 x4 + . . . for |x| < 1
2!
3!
22