Math 311 - hw 1 solutions
Tues, Sept 2014
1. Prove that n1/n → 1. (Hint: First show limx→∞ x1/x = 1).
We first use l’Hôpital’s rule (see 4.4.3) to show that limx→∞ lnxx = 0. Since the functions x, ln x are both differentiable on (1, ∞) (also both ln x, 1/x 6= 0 on (1, ∞)) and limx→∞ x = ∞ = limx→∞ ln x , l’Hôpital’s rule of the
form ∞/∞ applies. Hence we have
∞
1/x
ln x ∞
= lim
lim
= 0.
x→∞ 1
x→∞ x
Since the exponential function is continuous at 0 we have
ln x = exp 0 = 1.
lim x1/x = lim exp
x→∞
x→∞
x
Thus we have limn→∞ n1/n = limx→∞ x1/x = 1.
P∞
2. P
Suppose that
n=1 an is absolutely convergent and that {bn }n is a bounded sequence. Prove that the series
∞
n=1 an bn is absolutely convergent.
Since {bn }n is bounded, there is M > 0 such that |bn | ≤ M P
for all n ∈ N. Therefore we have |an bn | ≤PM |an | for
∞
∞
all n ∈ N. It follows by the Comparison test (see 6.1.9) that n=1 an bn is absolutely convergent, since n=1 an is
absolutely convergent.
3. In each case determine whether the given series is absolutely convergent, conditionally convergent or divergent.
Prove your answer, be sure to cite any tests you use by name whereever possible.
∞
X
(ln n)2
a.
(−1)n
n
n=10
We apply the Alternating Series Test (see 6.3.2) with an :=
(ln x)2
x
2
(ln n)2
n
to show that the series converges. We have
an ≥ 0 for all n ≥ 10. Now let f (x) =
for x ≥ 10; next we show that limx→∞ f (x) by arguing as in
exercise 1 above. Since the functions x, (ln x) are both differentiable on (1, ∞) (also (ln x)2 , 2 ln x/x 6= 0 on
(1, ∞)) and limx→∞ x = ∞ = limx→∞ (ln x)2 , l’Hôpital’s rule of the form ∞/∞ (see 4.4.3) applies and so
ln x
(ln x)2 ∞
∞
=
2 lim
= 0,
x→∞ x
x→∞
x
where the last equality follows from the application of l’Hôpital’s rule in exercise 1. Hence, an = f (n) → 0.
Next we show that {an }∞
n=10 is a decreasing sequence. By the quotient rule we have
lim
(2 − ln x) ln x
<0
for x > e2 .
x2
It follows that f is decreasing on (e2 , ∞) and so an = f (n) ≥ f (n + 1) = an+1 for all n ≥ 10. Hence the
Alternating Series Test applies and
P∞so the given series converges. But it is not absolutely convergent: we apply
the Integral Test to show that n=10 an diverges. We have shown above that f is a nonincreasing function
on [10, ∞); it is also nonnegative and continuous on [10, ∞) (and so integrable on any interval of the form
[10, T ]). Thus the Integral Test applies (see 6.2.1). Observe that (ln x)3 /3 is an antiderivative of f and so
T
Z ∞
Z T
1
1
(ln x)2
3
dx = lim (ln x) =
lim (ln T )3 − (ln 10)3 = ∞.
f (x) dx = lim
T
→∞
T
→∞
x
3
3 T →∞
10
10
10
P∞
Hence, n=10 an diverges and so the given series is conditionally convergent.
f 0 (x) =
b.
c.
∞
X
(−5)n n5
n!
n=1
We apply the Ratio Test (see 6.2.6) with an = (−5)n n5 /n!. We have
an+1 5n+1 (n + 1)5 n!
5(n + 1)5
5 1 5
=
=
=
1
+
→ 0.
an (n + 1)! 5n n5
(n + 1)n5
n+1
n
Since r = limn→∞ aan+1
= 0 < 1, the series is absolutely convergent.
n
∞
X
n
n + 3)n
((−1)
n=1
We apply the Root Test (see 6.2.4). Set an := ((−1)nn +3)n and observe that an =
if n is odd. Thus we have
(
n1/n /2 if n is even,
1/n
|an |
=
n1/n /4 if n is odd.
n
2n
if n is even and an =
Since n1/n → 1, we have ρ = lim supn |an |1/n = 1/2 < 1 and hence the series is absolutely convergent.
n
4n