Условие задачи [ править ] Найти сумму ряда:
∑
n
=
1
∞
n
⋅
sin
2
+
(
−
1
)
n
n
3
{\displaystyle \sum \limits _{n=1}^{\infty }n\cdot \sin {\frac {2+(-1)^{n}}{n^{3}}}}
Так как
2
+
(
−
1
)
n
⩽
3
,
{\displaystyle 2+(-1)^{n}\leqslant 3,}
2
+
(
−
1
)
n
n
3
⩽
3
n
3
(
1
)
{\displaystyle {\frac {2+(-1)^{n}}{n^{3}}}\leqslant {\frac {3}{n^{3}}}(1)}
∀
n
⩾
2
{\displaystyle \forall n\geqslant 2}
2
+
(
−
1
)
n
n
3
{\displaystyle {\frac {2+(-1)^{n}}{n^{3}}}}
3
n
3
{\displaystyle {\frac {3}{n^{3}}}}
(
0
;
π
2
)
,
{\displaystyle \left(0;{\frac {\pi }{2}}\right),}
sin
x
.
{\displaystyle \sin x.}
(
1
)
⇒
sin
2
+
(
−
1
)
n
n
3
⩽
sin
3
n
3
;
n
sin
2
+
(
−
1
)
n
n
3
⩽
n
sin
3
n
3
.
{\displaystyle (1)\Rightarrow \sin {\frac {2+(-1)^{n}}{n^{3}}}\leqslant \sin {\frac {3}{n^{3}}};n\sin {\frac {2+(-1)^{n}}{n^{3}}}\leqslant n\sin {\frac {3}{n^{3}}}.}
∑
n
=
1
∞
n
⋅
sin
3
n
3
{\displaystyle \sum \limits _{n=1}^{\infty }n\cdot \sin {\frac {3}{n^{3}}}}
∑
n
=
1
∞
1
n
2
,
{\displaystyle \sum \limits _{n=1}^{\infty }{\frac {1}{n^{2}}},}
lim
n
→
∞
n
sin
3
n
3
1
n
2
=
lim
n
→
∞
n
3
sin
3
n
3
=
lim
n
→
∞
n
3
3
n
3
=
3.
{\displaystyle \lim _{n\to \infty }{\frac {n\sin {\frac {3}{n^{3}}}}{\frac {1}{n^{2}}}}=\lim _{n\to \infty }n^{3}\sin {\frac {3}{n^{3}}}=\lim _{n\to \infty }n^{3}{\frac {3}{n^{3}}}=3.}
3
≠
0
{\displaystyle 3\neq 0}
3
≠
∞
,
{\displaystyle 3\neq \infty ,}
∑
n
=
1
∞
1
n
2
{\displaystyle \sum \limits _{n=1}^{\infty }{\frac {1}{n^{2}}}}
α
=
2
{\displaystyle \alpha =2}
∑
n
=
1
∞
n
⋅
sin
3
n
3
{\displaystyle \sum \limits _{n=1}^{\infty }n\cdot \sin {\frac {3}{n^{3}}}}
∑
n
=
1
∞
n
⋅
sin
2
+
(
−
1
)
n
n
3
.
{\displaystyle \sum \limits _{n=1}^{\infty }n\cdot \sin {\frac {2+(-1)^{n}}{n^{3}}}.}
