AP Calculus BC Quiz
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INFINITE SEQUENCES AND SERIES
Question 1
of 20
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Use the Integral Test to determine whether
∑
n
=
1
∞
1
(
n
+
1
)
2
\sum_{n=1}^{\infty}\frac{1}{(n+1)^2}
∑
n
=
1
∞
(
n
+
1
)
2
1
converges.
A
Diverges, because
∫
1
∞
1
(
x
+
1
)
2
d
x
\int_1^{\infty}\frac{1}{(x+1)^2}\,dx
∫
1
∞
(
x
+
1
)
2
1
d
x
diverges.
B
Converges, because
∫
1
∞
1
(
x
+
1
)
2
d
x
\int_1^{\infty}\frac{1}{(x+1)^2}\,dx
∫
1
∞
(
x
+
1
)
2
1
d
x
converges.
C
Diverges, because
∫
1
∞
1
(
x
+
1
)
2
d
x
=
ln
(
x
+
1
)
∣
1
∞
\int_1^{\infty}\frac{1}{(x+1)^2}\,dx=\left.\ln(x+1)\right|_1^{\infty}
∫
1
∞
(
x
+
1
)
2
1
d
x
=
ln
(
x
+
1
)
∣
1
∞
is infinite.
D
Converges, because
∫
1
∞
1
(
x
+
1
)
2
d
x
=
1
x
+
1
∣
1
∞
\int_1^{\infty}\frac{1}{(x+1)^2}\,dx=\left.\frac{1}{x+1}\right|_1^{\infty}
∫
1
∞
(
x
+
1
)
2
1
d
x
=
x
+
1
1
1
∞
is finite.
E
Converges, because the Integral Test applies even if
f
(
x
)
=
1
(
x
+
1
)
2
f(x)=\frac{1}{(x+1)^2}
f
(
x
)
=
(
x
+
1
)
2
1
is not continuous.
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