AP Calculus BC Quiz
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INFINITE SEQUENCES AND SERIES
Question 1
of 20
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An alternating-free series is
∑
n
=
1
∞
1
n
2
+
ln
n
\sum_{n=1}^{\infty} \frac{1}{n^2 + \ln n}
∑
n
=
1
∞
n
2
+
l
n
n
1
. Which comparison correctly determines convergence?
A
Since
1
n
2
+
ln
n
≥
1
n
2
\frac{1}{n^2+\ln n}\ge \frac{1}{n^2}
n
2
+
l
n
n
1
≥
n
2
1
for
n
≥
1
n\ge1
n
≥
1
and
∑
1
n
2
\sum \frac{1}{n^2}
∑
n
2
1
converges, the series converges.
B
Since
1
n
2
+
ln
n
≤
1
n
\frac{1}{n^2+\ln n}\le \frac{1}{n}
n
2
+
l
n
n
1
≤
n
1
for
n
≥
1
n\ge1
n
≥
1
and
∑
1
n
\sum \frac{1}{n}
∑
n
1
diverges, the series converges.
C
Since
1
n
2
+
ln
n
≤
1
n
2
\frac{1}{n^2+\ln n}\le \frac{1}{n^2}
n
2
+
l
n
n
1
≤
n
2
1
for
n
≥
1
n\ge1
n
≥
1
and
∑
1
n
2
\sum \frac{1}{n^2}
∑
n
2
1
converges, the series converges.
D
Since
1
n
2
+
ln
n
≥
1
n
\frac{1}{n^2+\ln n}\ge \frac{1}{n}
n
2
+
l
n
n
1
≥
n
1
for
n
≥
1
n\ge1
n
≥
1
and
∑
1
n
\sum \frac{1}{n}
∑
n
1
diverges, the series converges.
E
Since
1
n
2
+
ln
n
≤
1
n
3
\frac{1}{n^2+\ln n}\le \frac{1}{n^3}
n
2
+
l
n
n
1
≤
n
3
1
for
n
≥
1
n\ge1
n
≥
1
and
∑
1
n
3
\sum \frac{1}{n^3}
∑
n
3
1
converges, the series diverges.
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