Limit bo'yicha taqqoslash alomati (testi)
Matematikada
limit bo'yicha taqqoslash alomati (testi) (taqqoslash alomatidan farqli o'laroq) cheksiz qatorlarning yaqinlashishini tekshirish usuli hisoblanadi.
Sharti
Aytaylik, bizda ikkita
Σ
n
a
n
{\displaystyle \Sigma _{n}a_{n}}
va
Σ
n
b
n
{\displaystyle \Sigma _{n}b_{n}}
qatorlar berilgan bo'lib, barcha
n
{\displaystyle n}
lar uchun
a
n
≥
0
,
b
n
>
0
{\displaystyle a_{n}\geq 0,b_{n}>0}
munosabat o'rinli bo'lsin. Agar
lim
n
→
∞
a
n
b
n
=
c
{\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}
munosabat
0
<
c
<
∞
{\displaystyle 0<c<\infty }
uchun o'rinli bo'lsa, u holda har ikkala qator yaqinlashadi yoki har ikkala qator uzoqlashadi.
Isbot
lim
n
→
∞
a
n
b
n
=
c
{\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}
bo'lganligi uchun barcha
ε
>
0
{\displaystyle \varepsilon >0}
sonlar uchun shunday
n
0
{\displaystyle n_{0}}
musbat butun son mavjudki, bunda barcha
n
≥
n
0
{\displaystyle n\geq n_{0}}
lar uchun
|
a
n
b
n
−
c
|
<
ε
{\displaystyle \left|{\frac {a_{n}}{b_{n}}}-c\right|<\varepsilon }
munosabat, yoki shunga ekvivalent ravishda
−
ε
<
a
n
b
n
−
c
<
ε
{\displaystyle -\varepsilon <{\frac {a_{n}}{b_{n}}}-c<\varepsilon }
c
−
ε
<
a
n
b
n
<
c
+
ε
{\displaystyle c-\varepsilon <{\frac {a_{n}}{b_{n}}}<c+\varepsilon }
(
c
−
ε
)
b
n
<
a
n
<
(
c
+
ε
)
b
n
{\displaystyle (c-\varepsilon )b_{n}<a_{n}<(c+\varepsilon )b_{n}}
munosabatlar o'rinli.
c
>
0
{\displaystyle c>0}
bo'lganligi uchun
ε
{\displaystyle \varepsilon }
ni shunday yetarlicha kichik tanlashimiz mumkinki, bunda
c
−
ε
{\displaystyle c-\varepsilon }
musbat bo'ladi. Shunday qilib,
b
n
<
1
c
−
ε
a
n
{\displaystyle b_{n}<{\frac {1}{c-\varepsilon }}a_{n}}
va taqqoslash alomatiga ko'ra, agar
∑
n
a
n
{\displaystyle \sum _{n}a_{n}}
yaqinlashsa, u holda
∑
n
b
n
{\displaystyle \sum _{n}b_{n}}
ham yaqinlashadi.
Xuddi shunday,
a
n
<
(
c
+
ε
)
b
n
{\displaystyle a_{n}<(c+\varepsilon )b_{n}}
munosabat o'rinli va taqqoslash testiga ko'ra
∑
n
a
n
{\displaystyle \sum _{n}a_{n}}
uzoqlashsa, u holda
∑
n
b
n
{\displaystyle \sum _{n}b_{n}}
ham uzoqlashadi.
Ya'ni, har ikkala qator yaqinlashadi yoki har ikkala qator uzoqlashadi.
Misol
∑
n
=
1
∞
1
n
2
+
2
n
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}+2n}}}
qatorni yaqinlashishga tekshiramiz. Buning uchun uni yaqinlashuvchi bo'lgan
∑
n
=
1
∞
1
n
2
=
π
2
6
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}={\frac {\pi ^{2}}{6}}}
qator bilan taqqoslaymiz.
lim
n
→
∞
1
n
2
+
2
n
n
2
1
=
1
>
0
{\displaystyle \lim _{n\to \infty }{\frac {1}{n^{2}+2n}}{\frac {n^{2}}{1}}=1>0}
ekanligidan, berilgan qatorning ham yaqinlashuvchi bo'lishi kelib chiqadi.
Bir tomonlama versiya: bir tomonlama taqqoslash alomati
Bir tomonlama taqqoslash alomati (testi) yuqori limitni qo'llagan holda keltirilishi mumkin. Barcha
n
{\displaystyle n}
lar uchun
a
n
,
b
n
≥
0
{\displaystyle a_{n},b_{n}\geq 0}
bo'lsin. U holda agar
lim sup
n
→
∞
a
n
b
n
=
c
{\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=c}
munosabat
0
≤
c
<
∞
{\displaystyle 0\leq c<\infty }
uchun o'rinli bo'lsa va
Σ
n
b
n
{\displaystyle \Sigma _{n}b_{n}}
qator yaqinlashuvchi bo'lsa, u holda zaruriy ravishda
Σ
n
a
n
{\displaystyle \Sigma _{n}a_{n}}
qator ham yaqinlashadi.
Misol
Barcha natural
n
{\displaystyle n}
lar uchun
a
n
=
1
−
(
−
1
)
n
n
2
{\displaystyle a_{n}={\frac {1-(-1)^{n}}{n^{2}}}}
va
b
n
=
1
n
2
{\displaystyle b_{n}={\frac {1}{n^{2}}}}
bo'lsin. Ushbu
lim
n
→
∞
a
n
b
n
=
lim
n
→
∞
(
1
−
(
−
1
)
n
)
{\displaystyle \lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\lim _{n\to \infty }(1-(-1)^{n})}
limit mavjud bo'lmaganligi uchun biz odatiy taqqoslash alomatini qo'llay olmaymiz. Ammo,
lim sup
n
→
∞
a
n
b
n
=
lim sup
n
→
∞
(
1
−
(
−
1
)
n
)
=
2
∈
[
0
,
∞
)
{\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\limsup _{n\to \infty }(1-(-1)^{n})=2\in [0,\infty )}
munosabatdan va
∑
n
=
1
∞
1
n
2
{\displaystyle \sum _{n=1}^{\infty }{\frac {1}{n^{2}}}}
ning yaqinlashuvchi ekanligidan, bir tomonlama taqqoslash alomatiga ko'ra
∑
n
=
1
∞
1
−
(
−
1
)
n
n
2
{\displaystyle \sum _{n=1}^{\infty }{\frac {1-(-1)^{n}}{n^{2}}}}
qator yaqinlashuvchi bo'ladi.
Bir tomonlama taqqoslash alomatining teskarisi
Barcha
n
{\displaystyle n}
lar uchun
a
n
,
b
n
≥
0
{\displaystyle a_{n},b_{n}\geq 0}
bo'lsin. Agar
Σ
n
a
n
{\displaystyle \Sigma _{n}a_{n}}
qator uzoqlashuvchi bo'lib,
Σ
n
b
n
{\displaystyle \Sigma _{n}b_{n}}
qator yaqinlashuvchi bo'lsa, u holda zaruriy ravishda
lim sup
n
→
∞
a
n
b
n
=
∞
{\displaystyle \limsup _{n\to \infty }{\frac {a_{n}}{b_{n}}}=\infty }
, qaysiki,
lim inf
n
→
∞
b
n
a
n
=
0
{\displaystyle \liminf _{n\to \infty }{\frac {b_{n}}{a_{n}}}=0}
munosabat o'rinli bo'ladi. Bu yerda asosiy mazmun, qaysidir ma'noda,
a
n
{\displaystyle a_{n}}
larning
b
n
{\displaystyle b_{n}}
lardan kattaroq ekanligidadir.
Misol
D
=
{
z
∈
C
:
|
z
|
<
1
}
{\displaystyle D=\{z\in \mathbb {C} :|z|<1\}}
birlik diskda analitik bo'lgan
f
(
z
)
=
∑
n
=
0
∞
a
n
z
n
{\displaystyle f(z)=\sum _{n=0}^{\infty }a_{n}z^{n}}
berilgan bo'lib, u chekli yuzali tasvirga (aksga) ega bo'lsin. Parseval formulasi bo'yicha
f
{\displaystyle f}
ning tasviri(aks)ning yuzasi
∑
n
=
1
∞
n
|
a
n
|
2
{\displaystyle \sum _{n=1}^{\infty }n|a_{n}|^{2}}
ga teng. Bundan tashqari,
∑
n
=
1
∞
1
/
n
{\displaystyle \sum _{n=1}^{\infty }1/n}
qator uzoqlashuvchi. Shuning uchun, taqqoslash alomatining teskarisi bo'yicha,
lim inf
n
→
∞
n
|
a
n
|
2
1
/
n
=
lim inf
n
→
∞
(
n
|
a
n
|
)
2
=
0
{\displaystyle \liminf _{n\to \infty }{\frac {n|a_{n}|^{2}}{1/n}}=\liminf _{n\to \infty }(n|a_{n}|)^{2}=0}
, qaysiki,
lim inf
n
→
∞
n
|
a
n
|
=
0
{\displaystyle \liminf _{n\to \infty }n|a_{n}|=0}
.
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