# NAMES: MATH 152 April 1, 2015 QUIZ 7 ```NAMES:
MATH 152
April 1, 2015
QUIZ 7
on the final answer, but also on the work leading up to it.
1. (3 points) Determine whether or not the series converges. Show support for your answers.
∞
X
1
2
n + n ln n
n=1
Solution: This is done using the direct comparison:
Note that n2 +n1 ln n &lt; n12 . So
∞
X
n=1
∞
X 1
1
&lt;
&lt;∞
n2 + n ln n n=1 n2
(by p-test)
and the series converges.
2. (3 points) Determine whether or not the series converges. Show support for your answers.
∞
X
1
n ln n
n=2
Solution: Using the integral test:
∞
Z ∞
dx
= ln(ln x) = ∞
x ln x
2
2
gives that the series diverges.
(let u = ln x)
NAMES:
MATH 152
April 1, 2015
3. (3 points) Determine whether or not the series converges. Show support for your answers.
∞
X
1
√
n
ln
n
+
n
n=2
P
1
Solution: This is done using the limit comparison test with ∞
n=2 n ln n :
1 √
n ln n+ n
lim
1
n→∞
n ln n
Thus since
P
1
n ln n
n ln n
√
n→∞ n ln n +
n
1
√
= lim
= 1 6= 0
n
n→∞ 1 +
n ln n
= lim
diverges (by 2), the limit comparison test gives that the series diverges.
```