5 Steps to a 5 AP Calculus BC 2019

Series 351

Example 3

Determine whether the series 1+

1

2

+

1

3

+

1

4

+···+

1

n

+···=

∑∞

n= 1

1

n

converges or diverges.

Step 1: f(x)=

1

x

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

Step 2:

∫∞

1

1

x

dx=ulim→∞

∫u

1

1

x

dx=ulim→∞(lnx)

∣∣u

1 =ulim→∞[lnu−ln 1]=∞. Since the im-

proper integral does not converge, the series

∑∞

n= 1

1

n

diverges.

Example 4

Determine whether the series 1+

1

√

2

+

1

√

3

+

1

√

4

+···+

1

√

n

+···=

∑∞

n= 1

1

√

n

converges or diverges.

Step 1: f(x)=

1

√

x

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

Step 2:

∫∞

1

1

√

x

dx=ulim→∞

∫u

1

1

√

x

dx=ulim→∞

(

2

√

x

)

|u 1 =ulim→∞

[

2

√

u− 2

]

=∞

Since the improper integral does not converge, the series

∑∞

n= 1

1

√

n

diverges.

Ratio Test

If

∑∞

n= 1

anis a series with positive terms and limn→∞

an+ 1

an

<1, then the series converges. If the

limit is greater than 1 or is∞, the series diverges. If the limit is 1, another test must be used.

Example 1

Determine whether the series

∑∞

n= 1

(n+1)· 2 n

n!

converges or diverges.

Step 1: For alln≥1,
(n+1)· 2 n
n!
is positive.

Step 2: nlim→∞
an+ 1
an

= nlim→∞

(

(n+2)· 2 n+^1

(n+1)!

·

n!

(n+1)· 2 n

)

= nlim→∞

(

(n+2)· 2

(n+1)^2

)

= 0

Since this limit is less than 1, the series converges.

Example 2

Determine whether the series

∑∞

n= 1

n^3

(ln 2)n

converges or diverges.

Step 1: For alln≥1,
n^3
(ln 2)n

is positive.