MAC 2312 Quiz 6
February 28, 2012
SOLUTIONS
1.
Determine the nonzero values of
x
for which the series
∑
∞
n
=1
x
5
x
2
n

1
converges.
Solution:
First, observe that the numerator of the summand is independent of
n
, so
∞
summationdisplay
n
=1
x
5
x
2
n
−
1
=
x
5
∞
summationdisplay
n
=1
1
x
2
n
−
1
.
Therefore, we only need to find the values of
x
for which the sum on the righthand
side converges. To achieve this, consider two cases separately.
Case 1:

x

<
1
In this case, lim
n
→∞
x
2
n
= 0, so the limit of the summand is
lim
n
→∞
1
x
2
n
−
1
=
−
1
negationslash
= 0
.
Hence, by the test for divergence of series,
∑
∞
n
=1
1
x
2
n

1
is divergent.
Case 2:

x

>
1
In this case, lim
n
→∞
x
2
n
= +
∞
, so
lim
n
→∞
1
x
2
n
1
x
2
n

1
= lim
n
→∞
x
2
n
−
1
x
2
n
= 1
.
Therefore, by the limit comparison test,
∑
∞
n
=1
1
x
2
n

1
converges if and only if
∑
∞
n
=1
1
x
2
n
converges. But since

x

>
1,
1
x
2
<
1. Therefore in view of the geometric series,
∞
summationdisplay
n
=1
1
x
2
n
=
∞
summationdisplay
n
=1
parenleftbigg
1
x
2
parenrightbigg
n
<
∞
,
so
∑
∞
n

1
1
x
2
n

1
converges for

x

>
1.
2.
Determine the positive values of
k
for which the series
∑
∞
n
=2
√
n
n

n
k
converges.
Solution:
Here is the intuition behind finding the solution (this is not a solution; it
is just a way to come up with a candidate for a solution). If
k >
1, then as
n
→ ∞
,
n
k
dominates
n
, so the summand is approximately
√
n
n
−
n
k
∼
√
n
n
k
=
1
n
k

1
/
2
.
Therefore, by the
p
series test, we suspect that the given sum will converge if and
only if
k >
3
/
2 (because this makes
k
−
1
2
>
1).
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With a candidate for a solution in hand (i.e.
k >
3
2
), we are ready to check whether
our proposed solution is an actual solution. This amounts to performing two com
putations. First, we must show that the given series converges if
k >
3
2
. Second, we
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 Spring '08
 Bonner
 Calculus, lim

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