Since
a

1
< a
n
and
x
n
is the inf of
a
n
, a

1
≤
x
n
Since
x
n
≤
a
n
< a
n
+
1
by transitivity
x
n
<
1
Therefore

x
n

a
 ≤
1
=
2
<
So by transitivity

x
n

a

<
Since
wasarbitrary
∀
> o,
∃
N
∈
N
such that
∀
n
≥
N,

y
n

a

<
and

x
n

a

<
Therefore
y
n
and
x
n
converge to
a
by definition of converges.
So if lim
a
n
exists then lim inf
a
n
= lim sup
a
n
Suppose lim inf
a
n
= lim sup
a
n
Since
x
n
≤
a
n
≤
y
n
, by the Squeeze Theorem lim
x
n
= lim
a
n
= lim
y
n
Therefore if lim inf
a
n
= lim sup
a
n
then lim
a
n
exists
So we have if lim inf
a
n
= lim sup
a
n
then lim
a
n
exists and if lim
a
n
exists
then lim inf
a
n
= lim sup
a
n
Therefore lim inf
a
n
= lim sup
a
n
iff lim
a
n
exists
5
4
Find an explicit formula for the sequence and
determine if the series converges
∞
X
n
=1
1
n
(
n
+ 1)
∞
X
n
=1
1
n
(
n
+1)
=
∞
X
n
=1
1
n

1
n
+1
∞
X
n
=1
1
n

1
n
+1
= 1

1
2
+
1
2

1
3
+
1
3
...

1
k
+
1
k

1
k
+1
1

1
2
+
1
2

1
3
+
1
3
...

1
k
+
1
k

1
k
+1
= 1

1
k
+1
lim
k
→∞
1

1
k
+1
= 1
Therefore the series converges to 1
6
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 Fall '19
 lim, Mathematical analysis, Limit of a sequence