0.01 x 0.01 =
0 1
O
0.1
o
0.0001
0.001
Answers
Answer:
0.0001 is the answer hope it helps
Answer:
Answer:
A geometric series of geometric sequence
u
n
=
u
1
⋅
r
n
−
1
converges only if the absolute value of the common factor
r
of the sequence is strictly inferior to
1
; in other words, if
|
r
|
<
1
.
Explanation:
The standard form of a geometric sequence is :
u
n
=
u
1
⋅
r
n
−
1
And a geometric series can be written in several forms :
+
∞
∑
n
=
1
u
n
=
+
∞
∑
n
=
1
u
1
⋅
r
n
−
1
=
u
1
+
∞
∑
n
=
1
r
n
−
1
=
u
1
⋅
lim
n
→
+
∞
(
r
1
−
1
+
r
2
−
1
+
r
3
−
1
+
...
+
r
n
−
1
)
Let
r
n
=
r
1
−
1
+
r
2
−
1
+
r
3
−
1
+
...
+
r
n
−
1
Let's calculate
r
n
−
r
⋅
r
n
:
r
n
−
r
⋅
r
n
=
r
1
−
1
−
r
2
−
1
+
r
2
−
1
−
r
3
−
1
+
r
3
−
1
+
...
−
r
n
−
1
+
r
n
−
1
−
r
n
=
r
1
−
1
−
r
n
r
n
(
1
−
r
)
=
r
1
−
1
−
r
n
=
1
−
r
n
r
n
=
1
−
r
n
1
−
r
Therefore, the geometric series can be written as :
u
1
+
∞
∑
n
=
1
r
n
−
1
=
u
1
⋅
lim
n
→
+
∞
(
1
−
r
n
1
−
r
)
Thus, the geometric series converges only if the series
+
∞
∑
n
=
1
r
n
−
1
converges; in other words, if
lim
n
→
+
∞
(
1
−
r
n
1
−
r
)
exists.
If |r| > 1 :
lim
n
→
+
∞
(
1
−
r
n
1
−
r
)
=
∞
If |r| < 1 :
lim
n
→
+
∞
(
1
−
r
n
1
−
r
)
=
1
1
−
r
.
Therefore, the geometric series of geometric sequence
u
n
converges only if the absolute value of the common factor
r
of the sequence is strictly inferior to
1
.
Shura · 2 · Jul 2 2015
What are some examples of infinite geometric series?