Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky 1.3 Convergence Tests + Estimating Sums (8.3-8.4)
When a sequence of partial sum {Sn } of a series has a simple formula it is easy to find the sum, however this is not always the case. We will learn several convergence tests that doesn’t require evaluating a sum. In some cases these methods lead to estimating of the sum 1.3.1 The Integral test
Theorem: Let f be continues, positive decreasing function on [ m,∞ ) and let an = f ( n ) for n ≥ m ∞
∞
n=m
m
then the series ∑ an converges if and only if ∫ f ( x ) dx . Note: Since finite number of terms cannot affect convergence of infinite series, it is enough that f ( x ) decreasing on [ M ,∞ ) where M > m . The following graphical examples intuitively explain the theorem. 1
-­‐ The graph of the area under 2 is a convergent integral. This area bound the
x
series
1
π2
;
therefore
convergent
(to
, the proof is out of our scope).
∑ n2
6
n=1
∞
-­‐ The area under
∞
1
is infinite, i.e. divergent improper integral; therefore the series
x
∞
1
1
∑ n > ∫ x dx are also divergent.
n=1
1
Ex 1.
∞
Determine whether the series ∑ ne− n converges or diverges
2
n=1
Since we want to use integral test, we first show that f ( x ) = xe− x is positive and decreasing on [1,∞ ] . The function is trivially positive, since both x and exponent are positive on [1,∞ ] . To show it is decreasing function we will look at derivative 2
2
2
1
1
f ' ( x ) = e− x − 2x 2 e− x = (1− 2x 2 ) e− x = 0 ⇔ 1− 2x 2 = 0 ⇔ x 2 = ⇔ x = ±
, neither in our scope, 2
2
so no critical points in [1,∞ ] , i.e. the first derivative doesn’t change the sign in [1,∞ ] . Therefore, we can check the sign of the first derivative at any x > 1 , e.g. f ( 2 ) < 0 . Thus, we can proceed with the integral test. 2
6 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky ∞
λ
∞
2
1
( )
2
2
1 d − x2
1
d
1
e dx = − lim ∫ e− x dx = − lim e− x
2 dx
2 λ→∞ 1 dx
2 λ→∞
−x
∫ xe dx = ∫ −
1
∞
λ
1
(
)
2
1
1
== − lim e− λ − e−1 =
2 λ→∞
2e
Therefore ∑ ne− n converges. 2
n=1
Ex 2.
For what values of p is the series
∞
1
∑n
p
convergent?
n=1
∞
1
1
= 1+ 2|p| + 3|p| + ... = ∞ , another method lim p = ∞ ≠ 0 , any way it diverges. p
n→∞
n
n=1 n
For p<0: ∑
∞
1
1
For p=0: ∑ p = 1+ 1+ 1+ ... = ∞ , another method lim p = 1 ≠ 0 , anyway it diverges. n→∞
n
n=1 n
1
If p>0, then f ( x ) = p is positive and decreasing on [1,∞ ] . x
⎧
⎛ 1 ⎞
p >1
⎪convergent ⎜
∞
− p+1 t
− p+1
⎝ 1− p ⎟⎠
1
x
t
−1
1
⎪
1− p
= lim
=
lim ( t − 1) = ⎨
∫1 x p dx = lim
t→∞ − p + 1
t→∞
p −1
p − 1 t→∞
1
⎪divergent ⎛ 1 ⎞
0 < p ≤1
⎜⎝
⎟
⎪⎩
"∞" ⎠
1.3.2 Comparison Tests
1.3.2.1 The Comparison Test Thm: Let ∑ an , ∑ bn be series with an ,bn ≥ 0 , then -­‐ If
-­‐ If
∑b
∑b
∑ a is also convergent
then ∑ a is also divergent
n
is convergent and an ≤ bn then
n
is divergent and an ≥ bn
∞
n
n
∞
Ex 3.
1
1
1
1
1
< n ⇒∑ n
< ∑ n = 1 , thus n
convergent
2 +1 2
2 +1
n=1 2 + 1
n=1 2
Ex 4.
∞
2n − 1
2n
2
2n 2 ∞ 1 2 π 2 π 2
2n − 1
<
=
⇒
< ∑ 2=
=
, thus
is
∑
3
2
3
2
3
3
7n + 5n + 2 7n
7n
7 n=1 n
7 6
21
7n + 5n 2 + 2
n=1 7n
n
convergent
Ex 5.
1
n 1
1
=
≥ , thus
is divergent.
n
n
n
n
1.3.2.2 The Limit Comparison Test an
= c where c > 0 is a finite constant, then n→∞ b
n
Thm: Let ∑ an , ∑ bn be series with an ,bn ≥ 0 . If lim
either both series converge or both diverge. 7 Course: Accelerated Engineering Calculus II Instructor: Michael Medvinsky Ex 6.
Determine whether
∞
∞
∑a = ∑ n
n
n=1
∞
∞
1
∑b = ∑ n
n
n=1
n=1
n2
is converges or diverges. We choose
3
+1
for a limit comparison test:
n=1
n2
3
an
n3
1
n
+
1
lim = lim
= lim 3
= lim
=1
n→∞ b
n→∞ 1 / n
n→∞ n + 1
n→∞ 1+ 1 / n 3
n
∞
n2
is divergent. 3
n=1 n + 1
Thus, we got ∑
1.3.3 Alternating Series
∞
∞
∑ b = ∑ ( −1)
n
Def: Let an > 0 , then n=1
n=1
∞
∞
∑ c = ∑ ( −1)
n
n=1
n
an = −a1 + a2 − a3 + a4 − +...
are alternating series. n−1
an = a1 − a2 + a3 − a4 + −...
n=1
∞
∞
n=1
n=1
Thm: If an alternating series, either ∑ bn = ∑ ( −1) an = −a1 + a2 − a3 + a4 − +... or ∞
∞
∑ c = ∑ ( −1)
n−1
n
n=1
n
an = a1 − a2 + a3 − a4 + −... , where an > 0 satisfy n=1
1) an+1 ≤ an and 2) lim an = 0 n→∞
Then the series are converges. Ex 7.
Test the following series for convergence
Consider f ( x ) =
1
, one found f ' ( x ) =
x ln x
decreasing function. Thus ak+1 =
1⎞
⎛
− ⎜ ln x + x ⋅ ⎟
⎝
x⎠
( x ln x )
2
=
∞
(−1)k+1
∑ k ln k
k=2
− ( ln x + 1)
( x ln x )2
< 0,∀x ≥ 2 that f is 1
1
1
= 0 , ≤
= ak . Furthermore lim =
k→∞
k ln k
( k + 1) ln ( k + 1) k ln k
∞
(−1)k+1
therefore the series ∑
converges. k=2 k ln k
∞
7n 2
7
7
7n 2
n
= lim
= .
Ex 8.
The series ∑ ( −1) 2
diverges since lim
2
n→∞ 3n − 4
n→∞
4
3n − 4
n=1
3− 2 3
n
8