5 Steps to a 5 AP Calculus BC 2019

350 STEP 4. Review the Knowledge You Need to Score High

14.3 Convergence Tests

Main Concepts:Divergence Test, Integral Test, Ratio Test, Comparison Test, Limit

Comparison Test

Divergence Test

Given a series

∑∞

n= 1

an, if limn→∞an/=0, then the series diverges.

Example

Determine whether the series

∑∞

n= 1

2 n

3 n+ 1

converges or diverges.

The series

∑∞

n= 1

2 n

3 n+ 1

=

1

2

+

4

7

+

3

5

+.... Applying the Divergence Test, you have limn→∞an=

nlim→∞

2 n

3 n+ 1

=

2

3

/=0.

Therefore, the series diverges.

Integral Test

Ifan=f(n) wherefis a continuous, positive, decreasing function on [c,∞), then the series

∑∞

n= 1

anis convergent if and only if the improper integral

∫∞

c

f(x)dxexists.

Example 1

Determine whether the series

∑∞

n= 1

1

n^2

sin

π

n

converges or diverges.

Step 1: f(x)=

1

x^2

sin

π

x

is continuous, positive, and decreasing on the interval [2,∞).

Step 2:

∫∞

2

1

x^2

sin

π

x

dx =

1

π

ulim→∞cos

(π

x

)∣∣

∣∣

u

2

=

1

π

ulim→∞

[

cos

(π

u

)

−cos

(

π

2

)]

=

1

π

ulim→∞

[

cos

(π

u

)]

=

1

π

. The improper integral exists, so

∑∞

n= 1

1

n^2

sin

π

n

converges.

Example 2

Determine whether the series 1+

1

4

+

1

9

+

1

16

+···+

1

n^2

+···=

∑∞

n= 1

1

n^2

converges or diverges.

Step 1: f(x)=

1

x^2

is continuous, positive, and decreasing on the interval [1,∞).

Step 2:

∫∞

1

1

x^2

dx=ulim→∞

∫u

1

1

x^2

dx=ulim→∞(−x−^1 )

∣∣

∣

∣

u

1

=ulim→∞

[

− 1

u

−

− 1

1

]

=

ulim→∞

[

1 −

1

u

]

= 1

Since the improper integral exists, the series

∑∞

n= 1

1

n^2

converges.