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.