Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 1.3.4 Absolute convergence
Def: A series ∑ an is called absolutely convergent if the series of absolute values ∑ an is convergent. Thm: If a series ∑ an is absolutely convergent then it is convergent. It is true because 1) 0 ≤ an + an ≤ 2 an , 2) ∑ an is convergent and so 2∑ an and by comparison test ( ∑ ( an + an ) ≤ 2∑ an ) also ∑ an + an is convergent. Finally ∑ a = ∑(a
n
n
+ an ) − ∑ an Def: A series ∑ an is called conditionally convergent if ∑ an converges but ∑ an diverges. ∞
∑
Ex 1.
n=1
∞
∑
n=1
( −1)n−1
n
Ex 2.
2
( −1)n−1 = 1−
n
2
1 1 1
1 1
1
1
+ 2 − 2 + 2 − 2 + 2 − 2 ... is absolutely convergent since
2
2
3 4
5 6
7 8
∞
1
is convergent.
2
n=1 n
=∑
( −1)n−1 = 1− 1 + 1 − 1 + 1 − 1 ... is convergent due to alternating series
∑
∞
n=1
n
theorem: 1) bn+1 =
2
3
4
5
6
1
1
1
< = bn and 2) lim bn = lim = 0
n→∞
n→∞ n
n +1 n
However, it is not absolutely convergent since
∞
∑
( −1)n−1
n=1
n
∞
1
is divergent.
n=1 n
=∑
1.3.5 The Ratio Test
The following test is very useful in determining absolute convergence. The ratio test theorem: a
-­‐ If lim n+1 = L < 1 then the series ∑ an is absolutely convergent (and therefore
n→∞ a
n
convergent)
-­‐ If lim
n→∞
an+1
a
= L > 1 (including lim n+1 = ∞ ) then the series
n→∞
an
an
-­‐ If lim
n→∞
an+1
= 1 then the ratio test inconclusive, that is we have to use another
an
method to determine convergence or divergence of
9 ∑a
n
∑a
n
.
divergent.
Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 1 / ( n + 1)
an+1
np
n ⎞
⎛
= lim
=
lim
⎟ → 1 , thus the ratio test is
p = ⎜ lim
p
n→∞
n→∞ n + 1
an
1/ n
( ) ⎝ n→∞ n + 1⎠
p
Ex 3.
lim
n→∞
p
inconclusive in case of p-series. Indeed we know that it may converge or diverge
depends of value of p, while the ratio test is independent of that value.
an+1 ( −1) ( n + 1) / 7 n+1
7 n ⎛ n + 1⎞
1 ⎛ 1⎞
1
=
=
= ⎜ 1+ ⎟ → < 0 , thus
⎟
n 5
n+1 ⎜
n
an
7 ⎝ n ⎠
7 ⎝ n⎠
7
( −1) n / 7
n+1
Ex 4.
5
5
5
n
∑ ( −1)
n5
7n
converges.
an+1
( n + 1)! ⋅ n n = n! ( n + 1) ⋅ n n = ⎛ n ⎞ = ⎛ n + 1⎞
=
⎜
⎟ ⎜
⎟
an
( n + 1)n+1 n! ( n + 1) ( n + 1)n n! ⎝ n + 1⎠ ⎝ n ⎠
n
Ex 5.
n!
∑ n n converges. Similarly,
thus
−n
−1
⎛⎛ 1⎞n⎞
1
= ⎜ ⎜ 1+ ⎟ ⎟ → < 1
e
⎝⎝ n⎠ ⎠
nn
a
∑ n! diverges, since an+1 → e > 1 .
n
1.3.6 Estimating Sums
1.3.6.1 Remainder Estimate for the Integral Test Thm: Let ak = f ( k ) , where f(x) is a continues, positive, decreasing function for x ≥ n (these are ∞
m
conditions for a integral test). Consider that S = ∑ an is convergent series. Let Sm = ∑ an be a n=1
partial sum and Rm = S − Sm =
∞
∑a
n
n=1
∞
is the remainder, then n=m+1
m
≤ ∫ f ( x ) . m
∞
1
?
n=1 n + 1
a) How good S3 estimates S = ∑
Ex 6.
∑n
2
1
1
1
1
1 1 1 5 + 2 +1 8
=
+
+
= + + =
=
+ 1 1+ 1 4 + 1 9 + 1 2 5 10
10
10
3
n=1
∫ f ( x) ≤ R
m+1
∞
2
By the theorem above
∞
dx
dx
π
t
R3 ≤ ∫ 2
= lim ∫ 2
= lim arctan x 3 = lim ( arctant − arctan 3) = − arctan 3 ≈ 0.321
t→∞
t→∞
t→∞
x +1
x +1
2
3
3
t
so the estimation isn’t that good.
b) Can we improve the estimation without calculating more terms?
According to estimating theorem
∞
∫
m+1
∞
∞
m
m+1
f ( x ) ≤ Rm = S − Sm ≤ ∫ f ( x ) ⇒ Sm +
so 1.04498 ≈
∫
∞
f ( x ) ≤ S ≤ Sm + ∫ f ( x )
m
∞
∞
8 π
8
8
8 π
+ − arctan 4 = + ∫ f ( x ) ≤ S ≤ + ∫ f ( x ) = + − arctan 3 ≈ 1.12175
10 2
10 m+1
10 m
10 2
10 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky One can take a midpoint of this approximation, to get S ≈ 1.08336 . The more
precise sum is S = 1.07667 , so it is clear that we improved the approximation (for
Rm ≈ −6.69 × 10 −3 ).
c) How many terms do we need in order to make Rn < 10 −3 .
∞
dx
dx
π
t
Rm ≤ ∫ 2
= lim ∫ 2
= lim arctan x m = lim ( arctant − arctan m ) = − arctan m
t→∞
t→∞
t→∞
x +1
x +1
2
m
m
t
so we look for m such that
π
π
⎛π
⎞
− arctan m < 10 −3 ⇒ arctan m > − 10 −3 ⇒ m > tan ⎜ − 10 −3 ⎟ = 1000
⎝
⎠
2
2
2
1.3.6.2 Alternating Estimation Theorem Thm: If an alternating series S = ∑ bn = ∑ ( −1) an satisfy an+1 ≤ an and lim an = 0 then n
n→∞
Rm = S − Sm ≤ bn Ex 7.
How many terms are needed to approximate
approximation is accurate to within 10 −3 . Answer
11 ∑ ( −1)
n
1
so that the
n3
1
< 10 −3 ⇒ n 3 > 10 3 ⇒ n > 10
n3