10.
n
2
n
{\displaystyle {\frac {n}{2^{n}}}}
1
2
,
2
4
,
3
8
,
4
16
{\displaystyle {\frac {1}{2}},\,{\frac {2}{4}},\,{\frac {3}{8}},\,{\frac {4}{16}}}
11.
(
log
a
)
n
x
n
n
!
{\displaystyle {\frac {\left(\log a\right)^{n}x^{n}}{n!}}}
log
a
⋅
x
1
,
log
2
a
⋅
x
2
2
,
log
3
a
⋅
x
3
6
,
log
4
a
⋅
x
4
24
{\displaystyle {\frac {\log a\cdot x}{1}},\,{\frac {\log ^{2}a\cdot x^{2}}{2}},\,{\frac {\log ^{3}a\cdot x^{3}}{6}},\,{\frac {\log ^{4}a\cdot x^{4}}{24}}}
12.
(
−
1
)
n
−
1
x
2
n
−
2
(
2
n
−
1
)
!
{\displaystyle {\frac {\left(-1\right)^{n-1}x^{2n-2}}{\left(2n-1\right)!}}}
1
1
,
−
x
2
3
!
,
x
4
5
!
,
−
x
6
7
!
{\displaystyle {\frac {1}{1}},\,-{\frac {x^{2}}{3!}},\,{\frac {x^{4}}{5!}},\,-{\frac {x^{6}}{7!}}}
135. Infinite series.
n
{\displaystyle n}
(A )
1
,
1
2
,
1
4
,
1
8
,
⋯
,
1
2
n
−
1
{\displaystyle 1,\,{\frac {1}{2}},\,{\frac {1}{4}},\,{\frac {1}{8}},\,\cdots ,\,{\frac {1}{2^{n-1}}}}
and let
S
n
{\displaystyle S_{n}}
(B )
S
n
=
1
+
1
2
+
1
4
+
1
8
+
⋯
+
1
2
n
−
1
{\displaystyle S_{n}=1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+\cdots +{\frac {1}{2^{n-1}}}}
Evidently
S
n
{\displaystyle S_{n}}
n
{\displaystyle n}
when
n
=
1
,
S
1
=
1
=
1
,
when
n
=
2
,
S
2
=
1
+
1
2
=
1
1
2
,
when
n
=
3
,
S
3
=
1
+
1
2
+
1
4
=
1
3
4
,
when
n
=
4
,
S
4
=
1
+
1
2
+
1
4
+
1
8
=
1
7
8
,
⋯
when
n
=
n
,
S
n
=
1
+
1
2
+
1
4
+
1
8
+
⋯
+
1
2
n
−
1
=
2
−
1
2
n
−
1
,
{\displaystyle {\begin{array}{lcllclcl}{\text{when}}\ n&=&1,&S_{1}&=&1&=&1,\\{\text{when}}\ n&=&2,&S_{2}&=&1+{\frac {1}{2}}&=&1{\frac {1}{2}},\\{\text{when}}\ n&=&3,&S_{3}&=&1+{\frac {1}{2}}+{\frac {1}{4}}&=&1{\frac {3}{4}},\\{\text{when}}\ n&=&4,&S_{4}&=&1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}&=&1{\frac {7}{8}},\\&&&&&\cdots &&\\{\text{when}}\ n&=&n,&S_{n}&=&1+{\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+\cdots +{\frac {1}{2^{n-1}}}&=&2-{\frac {1}{2^{n-1}}},\end{array}}}
[ 1]
Mark off points on a straight line whose distances from a fixed point correspond to these different sums. It is seen that the point
corresponding to any sum bisects the distance between the preceding point and 2. Hence it appears geometrically that when
n
{\displaystyle n}
lim
S
n
=
2
{\displaystyle \lim S_{n}=2}
.
lim
n
→
∞
S
n
=
l
i
m
n
→
∞
(
2
−
1
2
n
−
1
)
=
2
{\displaystyle \lim _{n\to \infty }S_{n}=lim_{n\to \infty }\left(2-{\frac {1}{2^{n-1}}}\right)=2}
.
[ 2]
[Since when
n
{\displaystyle n}
increases without limit
1
2
n
−
1
{\displaystyle {\tfrac {1}{2^{n-1}}}}
approaches zero as a limit.]
↑ Found by 6, § 1 , for the sum of a geometric series
↑ Such a result is sometimes for the sake of brevity, called the sum of the series; but the student must not forget that 2 is not the sum but the limit of the sum , as the number of terms increases without limit.
