Math 152 Class Notes October 25, 2015 10.2 Series

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Math 152 Class Notes
October 25, 2015
10.2 Series
A series is the sum of the terms of a sequence {an }∞
n=1 :
a1 + a2 + a3 + · · · + an + · · ·
which is denoted, for short, by
∞
X
an
or
X
an
n=1
Example 1.
∞
P
n = 1 + 2 + 3 + ··· + n + ···
n=1
∞ (−1)n
P
1 1
(−1)n
Example 2.
= −1 + − + · · · +
+ ···
n
2 3
n
n=1
Denition. Given a series
∞
P
an = a1 + a2 + a3 + · · · + an + · · · ,
n=1
sn =
n
X
ai = a1 + a2 + a3 + · · · + an
i=1
is called the n-th partial sum of the series an .
If the sequence {sn } is convergent and lim sn = s exists as a real number, then the
n→∞
P
series an is called convergent and we write
P
a1 + a2 + a3 + · · · + an + · · · = s
or
∞
X
an = s
n=1
The number is called the sum of the series. If the sequence {sn } is divergent, then
the series is called divergent.
Example 3. Determine whether the series
∞
P
(−1)n is convergent or divergent.
n=1
Example 4. Find the sum of
∞ 1
P
n
n=1 2
With any series
∞
P
an we associate two sequences: the sequence of its partial sums
n=1
{sn } and the sequence of its terms {an }.
sn =
n
X
ai = a1 + a2 + a3 + · · · + an ,
an = sn − sn−1
i=1
Example 5. The n-th partial sum of a series
∞
P
an is given by sn =
n=1
(a) Find the sum of the series or explain why the series is divergent.
(b) Find the sequence {an } and lim an .
n+1
.
2n + 3
n→∞
This example is a special case of the following theorem.
Theorem. If the series
∞
P
an is convergent, then lim an = 0.
n→∞
n=1
As corollary, we have the follow test for divergence.
Divergence test. If n→∞
lim an 6= 0, then the series
∞
P
an is divergent.
n=1
∞ n+1
P
Example 6. Show that the series
is divergent.
n=1 2n + 3
Example 7. Show that the series
∞
P
n=1
cos nπ is divergent.
Note. The converse of the divergence test is not true in general. If n→∞
lim an = 0, we
cannot conclude that
∞
P
an is convergent. Therefore if you nd that lim an = 0, then
n=1
the test for divergence fails and thus another test must be applied.
Example 8. Consider an =
(a) Show lim an = 0.
n→∞
(b) Show that the series
√
∞
P
n=1
n+1−
√
n.
an is divergent.
n→∞
The previous series is an example of telescoping series that is a series of the form
∞
P
(an+i − an ) for some integer i ≥ 1. Because of all the cancellations, the sum col-
n=1
lapses into just few terms.
Example 9. Find the sum of
∞ 1
P
1
).
( −
n+1
n=1 n
Example 10. Find the sum of
∞
P
(2n − 2n−1 ).
n=0
Example 11. Find the sum of
∞
P
n=1
ln
n+1
.
n+2
Example 12. Find the sum of
Example 13. Find the sum of
∞ 1
P
1
( −
).
n+2
n=1 n
∞
P
1
.
n=1 n(n + 2)
Example 14. Geometric series
∞
P
arn−1 = a + ar + ar2 + · · · + arn−1 + · · · .
n=1
Starting with the rst term a, each term is obtained from the preceding one by multiplying it by the common ratio r.
Find the value of r for which
∞
P
n=1
arn−1 is convergent and nd the sum.
Example 15. Find the sum of the series
∞
P
xn where |x| < 1.
n=0
Example 16. Find the sum of the geometric series
5−
10 20 40
+
−
+ ···
3
9
27
∞ 22n
P
Example 17. Is the series
convergent or divergent?
1−n
n=1 3
Example 18. Write 0.27 = 0.27272727 . . . as
(a) an innite series
(b) a fraction (ratio of integers)
Properties of convergent series
∞
P
If
an and
n=1
∞
P
constant),
(i)
(ii)
(iii)
∞
P
bn are convergent series, then so are the series
n=1
(an + bn ), and
n=1
∞
P
can = c
n=1
∞
P
∞
P
n=1
(an + bn ) =
n=1
(an − bn ) and
n=1
an
∞
P
n=1
∞
P
n=1
∞
P
n=1
n=1
(an − bn ) =
∞
P
an +
an −
∞
P
∞
P
n=1
∞
P
bn
bn
n=1
∞ 5
P
3
).
( n+
Example 19. Find the sum of
n(n + 1)
n=1 2
can (where c is a
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