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 ${\mathrm{\Sigma }}_n{a}_n$ va ${\mathrm{\Sigma }}_n{b}_n$ qatorlar berilgan bo'lib, barcha $n$ lar uchun $a_n\ge 0,b_n>0$ munosabat o'rinli bo'lsin. Agar $\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=c$ munosabat $0<c<\mathrm{\infty }$ uchun o'rinli bo'lsa, u holda har ikkala qator yaqinlashadi yoki har ikkala qator uzoqlashadi.

Isbot

$\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=c$ bo'lganligi uchun barcha $\epsilon >0$ sonlar uchun shunday $n_0$ musbat butun son mavjudki, bunda barcha $n\ge n_0$ lar uchun $|\frac{a_n}{b_n}-c|<\epsilon$ munosabat, yoki shunga ekvivalent ravishda

$-\epsilon <\frac{a_n}{b_n}-c<\epsilon$

$c-\epsilon <\frac{a_n}{b_n}<c+\epsilon$

$(c-\epsilon )b_n<a_n<(c+\epsilon )b_n$

munosabatlar o'rinli.

$c>0$ bo'lganligi uchun $\epsilon$ ni shunday yetarlicha kichik tanlashimiz mumkinki, bunda $c-\epsilon$ musbat bo'ladi. Shunday qilib, $b_n<\frac{1}{c-\epsilon }{a}_n$ va taqqoslash alomatiga ko'ra, agar $\sum _n{a}_n$ yaqinlashsa, u holda $\sum _n{b}_n$ ham yaqinlashadi.

Xuddi shunday, $a_n<(c+\epsilon )b_n$ munosabat o'rinli va taqqoslash testiga ko'ra$\sum _n{a}_n$ uzoqlashsa, u holda $\sum _n{b}_n$ ham uzoqlashadi.

Ya'ni, har ikkala qator yaqinlashadi yoki har ikkala qator uzoqlashadi.

Misol

$\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2+2n}$ qatorni yaqinlashishga tekshiramiz. Buning uchun uni yaqinlashuvchi bo'lgan $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$qator bilan taqqoslaymiz.

$\underset{n\to \mathrm{\infty }}{\lim }\frac{1}{n^2+2n}\frac{n^2}{1}=1>0$ 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 $n$ lar uchun $a_n,b_n\ge 0$ bo'lsin. U holda agar $\underset{n\to \mathrm{\infty }}{\mathrm{lim\; sup}}\frac{a_n}{b_n}=c$ munosabat $0\le c<\mathrm{\infty }$ uchun o'rinli bo'lsa va ${\mathrm{\Sigma }}_n{b}_n$ qator yaqinlashuvchi bo'lsa, u holda zaruriy ravishda ${\mathrm{\Sigma }}_n{a}_n$ qator ham yaqinlashadi.

Misol

Barcha natural $n$ lar uchun $a_n=\frac{1-(-1{)}^n}{n^2}$ va $b_n=\frac{1}{n^2}$ bo'lsin. Ushbu$\underset{n\to \mathrm{\infty }}{\lim }\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{\lim }(1-(-1{)}^n)$ limit mavjud bo'lmaganligi uchun biz odatiy taqqoslash alomatini qo'llay olmaymiz. Ammo, $\underset{n\to \mathrm{\infty }}{\mathrm{lim\; sup}}\frac{a_n}{b_n}=\underset{n\to \mathrm{\infty }}{\mathrm{lim\; sup}}(1-(-1{)}^n)=2\in [0,\mathrm{\infty })$ munosabatdan va $\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n^2}$ ning yaqinlashuvchi ekanligidan, bir tomonlama taqqoslash alomatiga ko'ra $\sum _{n=1}^{\mathrm{\infty }}\frac{1-(-1{)}^n}{n^2}$ qator yaqinlashuvchi bo'ladi.

Bir tomonlama taqqoslash alomatining teskarisi

Barcha $n$ lar uchun $a_n,b_n\ge 0$ bo'lsin. Agar ${\mathrm{\Sigma }}_n{a}_n$ qator uzoqlashuvchi bo'lib, ${\mathrm{\Sigma }}_n{b}_n$ qator yaqinlashuvchi bo'lsa, u holda zaruriy ravishda $\underset{n\to \mathrm{\infty }}{\mathrm{lim\; sup}}\frac{a_n}{b_n}=\mathrm{\infty }$, qaysiki, $\underset{n\to \mathrm{\infty }}{\mathrm{lim\; inf}}\frac{b_n}{a_n}=0$ munosabat o'rinli bo'ladi. Bu yerda asosiy mazmun, qaysidir ma'noda, $a_n$ larning $b_n$ lardan kattaroq ekanligidadir.

Misol

$D=\{z\in ℂ:|z|<1\}$ birlik diskda analitik bo'lgan$f(z)=\sum _{n=0}^{\mathrm{\infty }}{a}_n{z}^n$ berilgan bo'lib, u chekli yuzali tasvirga (aksga) ega bo'lsin. Parseval formulasi bo'yicha $f$ ning tasviri(aks)ning yuzasi $\sum _{n=1}^{\mathrm{\infty }}n|a_n{|}^2$ ga teng. Bundan tashqari, $\sum _{n=1}^{\mathrm{\infty }}1/n$ qator uzoqlashuvchi. Shuning uchun, taqqoslash alomatining teskarisi bo'yicha,$\underset{n\to \mathrm{\infty }}{\mathrm{lim\; inf}}\frac{n|a_n{|}^2}{1/n}=\underset{n\to \mathrm{\infty }}{\mathrm{lim\; inf}}(n|a_n|{)}^2=0$, qaysiki, $\underset{n\to \mathrm{\infty }}{\mathrm{lim\; inf}}n|a_n|=0$.

Yana qarang

• Yaqinlashish alomatlari
• Taqqoslash alomati (testi)

Manbalar

1. Andoza:Manbalar

Qo'shimcha manbalar

• Rinaldo B. Schinazi: From Calculus to Analysis. Springer, 2011, Andoza:Isbn, pp. 50
• Michele Longo and Vincenzo Valori: The Comparison Test: Not Just for Nonnegative Series. Mathematics Magazine, Vol. 79, No. 3 (Jun., 2006), pp. 205–210 (JSTOR)
• J. Marshall Ash: The Limit Comparison Test Needs Positivity. Mathematics Magazine, Vol. 85, No. 5 (December 2012), pp. 374–375 (JSTOR)

Havolalar

• Pauls Online Notes on Comparison Test