Quiz6 Solutions - MAC 2312 Quiz 6 SOLUTIONS x 1 Determine...

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 right-hand
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).

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