HW #5 Solutions (Math 323)
!
!
!
! (n+1)4 ·2n
1
4 1
14.1) a) The Ratio Test gives ! sn+1
sn ! = n4 ·2n+1 = (1 + 1/n) · 2 → 2 < 1 so the series converges.
!
!
!
!
2n+1 ·n!
2
b) The Ratio Test gives ! sn+1
sn ! = 2n ·(n+1)! = n+1 → 0 < 1 so the series converges.
!
!
!
! (n+1)2 ·3n
1
2 1
c) The Ratio Test gives ! sn+1
sn ! = n2 ·3n+1 = (1 + 1/n) · 3 → 3 < 1 so the series converges.
" n!
d) First, the series is bounded below by
n4 /2 , which by the Ratio Test diverges. Hence, the series diverges.
" 1
e) Taking the absolute values of the terms, we see that the series is bounded above by
n2 , which converges.
Hence, the series is absolutely convergent, showing that it is convergent.
"1
f) This series is bounded below by
n , which diverges. Hence, the series diveges.
" 1
n−n/2
n−1
1
14.2) a) We have n2 > n2 = 2n . Since
2n diverges, by the Comparison Test the given series diverges.
b) The terms do not → 0. Thus, the series diverges.
" 3
c) This is
n2 , which we know converges.
d) This is just like 14.1c and converges by the Ratio Test.
e) The Ratio Test shows this converges.
f) The Root Test shows list converges since lim sup |sn |1/n = lim sup n1 = 0 < 1.
g) This series is bounded above by the series in 14.1a, and hence converges.
!
!
#
!
! √ √n!
1
14.3) a) The Ratio Test gives us ! sn+1
=
=
!
sn
n+1 → 0 < 1, so the series converges.
(n+1)!
1
b) The absolute value of the nth term is bounded above by 2+1
3n = 3n−1 , which gives a geometric series with
1
multiplicative parameter r = 3 < 1. Since such a geometric series converges, the given series converges absolutely,
and hence converges.
" 1
1
c) The nth term is bounded above by 21n . Since
2n is a geometric series with r = 2 < 1, it converges. Thus,
the given series converges.
"
" 1
d) This series is bounded by 52 21n . Since
2n converges, the given series converges.
e) The terms do not → 0, hence the series diverges.
!
!
!
!
100
f) Use the Ratio Test: ! sn+1
sn ! = n+1 → 0 < 1, so the series converges.
"
" 1
14.4) a) This is bounded above by 1 (n − n/2)2 = 4
n2 which we know converges. Hence, the series converges.
b)
Write
out
the
first
few
terms.
You
should
notice
that
the nth term in the sequence of partial sums is equal to
√
n + 1 − 1. Since this does not → 0, the series diverges.
!
!
$
%n $
%n
!
!
(n+1)!·nn
n+1
n
1
c) Use the Ratio Test: ! sn+1
=
=
·
=
1
−
→ e−1 < 1, so the series converges.
!
n+1
sn
n!(n+1)
n+1
n+1
n+1
"
"m
14.7) Since
an converges, given ! < 1, there exists N such that for all m, n > N we have
i=n an < !. Let
"
∞
! = 1. Then, for all n > N , an < 1. Since p > 1, we have apn < an for all n > N . Since, i=N an is finite (the
"∞
"N
series converges), we see that
apn is finite (Comparison Test). Finally, since i=1 apn is finite (it’s a finite
"∞ i=N
number of terms), we have i=1 apn is finite (it converges).
14.8) Use the hint, which requires us showing that an bn ≤ (an + bn )2 = a2n + b2n + 2an bn . Hence, we want to show
that a2n + b2n + an bn ≥ 0. This is true since all terms
" are nonnegative. Let ! > 0 be given. Then
" there exists
N1 and N2 such that for all n, m > N1 we have "
an < !/2 and for all n,√
m > N2 we have
bn < !/2. Let
N = max(N1 , N2 ). "
Then,
for
all
n,
m
>
N
we
have
(a
+
b
)
<
!.
Since,
a
b
≤
a
+
b
we
see
that for all
n
n
n
n
n
n
√
"√
n, m > N we have
an bn < !, showing that
an bn converges.
14.10)
(−1)n )n . Then lim sup$|sn |1/n
! Let
! sn = (2 +n+1
% = 3 > 1 so the series divegers. However, the $Ratio%Test gives
! sn+1 !
sn+1
(2+(−1)
)n+1
us ! sn ! = (2+(−1)n )n . Hence, lim sup sn
= ∞ > 1 (considering n odd) while lim inf sn+1
=0<1
sn
(considering n even), thereby showing that the Ratio Test gives no information.
14.12) Let ank be an element of (an ) such that |ank − 0| < k12 . Since lim inf |an | = 0 = lim inf an , this is possible
"
" 1
(if it weren’t there would be a k̂ such |an − 0| > k̂12 for all n, showing that lim inf |an | > 0). Since
|ank | ≤
k2
"
(which converges), by the comparison test
ank converges absolutely, and hence is a convergent series.
1
14.13) a) The first is a geometric series with a = 23 and r = 23 . Hence, the sum is a · 1−r
= 1. The second one has
2
2
2
a = − 3 and r = − 3 , giving a sum of − 5 .
1
b) Use the hint. We see that the nth partial sum (which is the sum in the hint) is equal to sn = 1 − n+1
→ 1. By
definition, the sum is 1.
1
c) Use the hint. We see that the nth partial sum is equal to sn = 12 − 2n+1
n+1 → 2 .
n+2
d) Use the method in"parts (b) and (c) and the fact that 2nn = 2n+1
We get that the nth partial sum is
n−1 − 2n
n+2
n
sn = 2 − 2n → 2 =
.
n
2
"1
14.14) Clearly n1 < 2"log1n#+1 (which is the sequence given). We may write the given sequence as
2 , which clearly
diverges.
15.1) a) Converges (alternates, terms → 0).
b) Diverges, terms do not → 0.
15.2) a) Note that for n = 3 + 6k, sin(nπ/6) = ±1. Hence, the terms do not → 0, so the series diverges.
"
b) Converges. Note that for all n, sin(nπ/7) < .99. Hence, the series is bounded above by
(.99)n , which
converges. Hence, by the Comparison test, the series converges.
1
15.3) Use the Integral test with f (x) = x(log
have
x) for x ≥ 2 (which is positive and decreasing on [2, ∞)). We
!∞
& ∞ dx
& ∞ dx
(log x)1−p !
∞
= log log x|2 = ∞ so the series diverges for p = 1. For p '= 1, we have 2 x(log x)p =
! . For
1−p
2 x(log x)
!
! 2
1−p !∞
1−p !∞
x)
(log x)
p < 1 we see that 1 − p > 0 so that (log1−p
! = ∞, while for p > 1, we have 1 − p < 0 so that
! < ∞.
1−p
2
2
By the Integral Test, the series diverges for p ≤ 1 and converges for p > 1.
"1
√
15.4) a) We know that n ≥ log n so the given series is bounded below by
n , which diverges. Hence, the given
series diverges.
"1
b) The given series is bounded below by
n and hence diverges.
&∞
1
c) Use the Integeral Test with f (x) = x log x log log x . For integrating 2 f (x) dx use u-substituion with u = log log x
&∞
so that du = log1 x · x1 dx to see that 2 f (x) dx = log log log x|∞
2 = ∞. Hence, the series diverges.
" 1
√
d) We know that log n < n so the series is bounded above by
, which converges. Hence, the given series
n3/2
converges.
" 1
"1
15.5)
np ≤
n = ∞. All this tells us is that
" 1 I assume what they are getting at is that you would have
≤
∞,
which
tells
us
absolutely
nothing.
p
n
"1
15.6) a)
b) Yes, we’ve done this.
c) From part (b), it must contain negative terms. By the alternating
n
" (−1)n
"1
√
series test we know that
converges.
Squaring
all terms gives
n , which diverges.
n
15.7) We know that
" nan ≤ an +an+1 +· · ·+a2n−1 since the sequence (an ) is nonincreasing and an > 0 for all n since
(an ) → 0 (since
an converges) and (an ) is nonincreasing. Hence, we will show that an + an+1 + · · · + a2n−1 → 0,
thereby showing that nan → 0. But by the Cauchy Criterion for convergent series, there exists N such that for
all n, m > N , an + an+1 + · · · + am < ! (we can drop the absolute value signs since everything is positive). Take
m = 2n − 1 and we are done.
"1
1
b) From part a), is we assume
n converges, then lim n · n = 0, which is clearly not the case.
2