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Solutions to PS #9
3.6 (a) Let (sn ) be the sequence of partial sums. Notice that
√
√
√
√
√
2 − 1 + 3 − 2 + 4 − 3 + ···
sn =
√
√
√
√
√
√
√
= −1 + ( 2 − 2) + ( 3 − 3) + · · · + ( n − n) + n + 1
√
n+1−1
=
→ +∞.
Thus, the series
P
an diverges.
(b) Here
√
√
√
1
1
n+1+ n
√
≤
.
an =
· √
√ =
√
2n3/2
n+1+ n
n( n + 1 + n)
P
P
Since n n−3/2 is a convergent p-series, n an converges (by the comparison test).
√
(c) First, note that for n > 1, n n > 1. Also, it may be proven (see Thm. 3.20(c))
√
√
that limn n n = 1. Thus, we may choose N ∈ N s.t. n ≥ N ⇒ 1 < n n < 3/2.
And so, for n ≥ N ,
n
∞
X
√
1
n
n
( n − 1) ≤
an converges,
⇒
2
n=N
n+1−
n
√
n
P
−n
since ∞
converges (it is the tail of a convergent geometric series). Thus
n=N 2
P
P∞
n=N an .
n an converges, since it includes at most finitely-many more terms than
Of course, the root test works very well here, too.
3.11 (a) We prove this result handling two cases.
Case: (an ) is bounded by M > 0 (i.e., an < M , ∀n).
Then
an
an
,
∀n .
>
1 + an
1+M
But,
X an
X
1
=
an ,
1
+
M
1
+
M
n
n
a constant times a divergent series (also divergent). By the comparison test,
P
n an /(1 + an ) diverges.
Case: (an ) is unbounded.
Choose an M ∈ R s.t. M/(1 + M ) > 1/2. (M = 2 will do.) Since (an ) is unbounded, an > M for infinitely many n ∈ N. Let us write S := {n ∈ N | an > M },
an infinite set. Using the curve-sketching ideas from first-semester calculus, one
can show that the function f (x) := x/(1 + x) is an increasing function on (0, ∞).
Hence
M
1
an
>
>
,
for n ∈ S .
1 + an
1+M
2
P
Since S is infinite, this implies that an /(1 + an ) 6→ 0, and n an /(1 + an ) diverges
by Theorem 3.23.
P
An alternate proof is one by contradiction: Suppose that n an /(1 + an ) < ∞.
Then Thm. 3.23 indicates that
1
an
= 1−
→ 0,
1 + an
1 + an
which implies 1/(1 + an ) → 1, and hence an → 0. So, ∃ N ∈ N s.t. an < 1, ∀ n ≥
N , and
an
an
an
>
=
,
for n ≥ N.
1 + an
1+1
2
P
Under our assumption that n an /(1 + an ) converges, the comparison test says
P
P
that ∞
n=N an converges (and, of course,
n an must as well, since it includes at
most finitely many more terms).
→←