Section 3.4
#4)
(a)
x
n
= 1

(

1)
n
+ 1
/n
.
We take the subsequence of the evenindexed terms:
x
n
k
=
x
2
k
=
1

(

1)
2
k
+ 1
/
(2
k
) = 2 + 1
/
(2
k
).
So (
x
2
k
) converges to 2.
Next we take the subsequence of
oddindexed terms:
x
2
k

1
= 1

(

1)
2
k

1
+ 1
/
(2
k

1) = 1
/
(2
k

1). So (
x
2
k

1
) converges to 0.
Since there are two subsequences converging to different limits, the entire sequence diverges.
(b)
x
n
= sin(
nπ/
4). We first take the subsequence of multiples of 8:
x
8
k
= sin(8
kπ/
4) = sin(2
kπ
) = 0.
So the subsequence (
x
8
k
) converges to 0.
Now we take the subsequence where
n
k
= 8
k
+ 2:
x
8
k
+2
= sin((8
k
+ 2)
π/
4) = sin(2
πk
+
π/
2) = sin(
π/
2) = 1.
So (
x
8
k
+2
) converges to 1.
Since
there are two subsequences converging to different limits, the entire sequence diverges.
#6)
(a)
x
n
+1
< x
n
⇐⇒
(
n
+ 1)
1
/
(
n
+1)
< n
1
/n
⇐⇒
(
n
+ 1)
n
< n
n
+1
⇐⇒
(
n
+ 1)
n
n
n
< n
⇐⇒
1 +
1
n
n
< n
It was shown in Example 3.3.6 that (1+1
/n
)
n
<
3. Hence, for
n
≥
3, we have (1+1
/n
)
n
<
3
≤
n
,
and therefore
x
n
+1
< x
n
for all
n
≥
3. Thus, the sequence is ultimately decreasing (its 3tail is
decreasing) and since it is bounded below by 1,
x
:= lim(
x
n
) exists. Also
x
≥
1.
(b) (
x
2
n
) also converges to
x
, and
x
2
n
= (2
n
)
1
/
(2
n
)
=
√
2
1
/n
√
n
1
/n
=
√
2
1
/n
√
x
n
. Taking the limit
of both sides and recalling that 2
1
/n
converges to 1, we arrive at
x
=
√
x
. Since
x
≥
1, the only
viable solution is
x
= 1.
#7)
(a)
(1 + 1
/n
2
)
n
2
is a subsequence of ((1 + 1
/N
)
N
), which we know converges to
e
. Therefore, the
sequence in question also converges to
e
.
(b)
(
(1 + 1
/
2
n
)
2
n
)
is a subsequence of ((1 + 1
/N
)
N
), which we know converges to
e
. Therefore, the
subsequence also converges to
e
. Hence, ((1 + 1
/
2
n
)
n
) =
(1 + 1
/
2
n
)
2
n
→
√
e
.
(c)
(1 + 1
/n
2
)
2
n
2
=
(1 + 1
/n
2
)
n
2
2
. So the sequence in question converges to
e
2
by part (a)
and the algebraic properties of limits.
(d) There is no way to manipulate (1+2
/n
)
n
to get a subsequence of (1+1
/N
)
N
. By problem 5 (proven
below), it suffices to show that the evenindexed and oddindexed subsequences (1 + 2
/
(2
k
))
2
k
and (1 + 2
/
(2
k

1))
2
k

1
converge to the same thing. Evenindexed sequence: (1 + 2
/
(2
k
))
2
k
=
(
1 + 1
/k
)
k
)
2
, which conveges to
e
2
. Oddindexed sequence: note that
(1 + 2
/
(2
k
))
2
k

1
<
(1 + 2
/
(2
k

1))
2
k

1
<
(1 + 2
/
(2
k

2))
2
k

1
.
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 Spring '08
 JUNGE
 Limit, Trigraph, Cauchy, Xn, subsequence, NK

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