Matematikada **limit bo'yicha taqqoslash alomati (testi)** (taqqoslash alomatidan farqli o'laroq) cheksiz qatorlarning yaqinlashishini tekshirish usuli hisoblanadi.

Sharti
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Aytaylik, bizda ikkita {\displaystyle \Sigma \_{n}a\_{n}} va {\displaystyle \Sigma \_{n}b\_{n}} qatorlar berilgan bo'lib, barcha {\displaystyle n} lar uchun {\displaystyle a\_{n}\geq 0,b\_{n}>0} munosabat o'rinli bo'lsin. Agar {\displaystyle \lim \_{n\to \infty }{\frac {a\_{n}}{b\_{n}}}=c} munosabat {\displaystyle 0<c<\infty } uchun o'rinli bo'lsa, u holda har ikkala qator yaqinlashadi yoki har ikkala qator uzoqlashadi.

Isbot
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{\displaystyle \lim \_{n\to \infty }{\frac {a\_{n}}{b\_{n}}}=c} bo'lganligi uchun barcha {\displaystyle \varepsilon >0} sonlar uchun shunday {\displaystyle n\_{0}} musbat butun son mavjudki, bunda barcha {\displaystyle n\geq n\_{0}} lar uchun {\displaystyle \left|{\frac {a\_{n}}{b\_{n}}}-c\right|<\varepsilon } munosabat, yoki shunga ekvivalent ravishda

{\displaystyle -\varepsilon <{\frac {a\_{n}}{b\_{n}}}-c<\varepsilon }
{\displaystyle c-\varepsilon <{\frac {a\_{n}}{b\_{n}}}<c+\varepsilon }
{\displaystyle (c-\varepsilon )b\_{n}<a\_{n}<(c+\varepsilon )b\_{n}}
munosabatlar o'rinli.

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

Xuddi shunday, {\displaystyle a\_{n}<(c+\varepsilon )b\_{n}} munosabat o'rinli va taqqoslash testiga ko'ra{\displaystyle \sum \_{n}a\_{n}} uzoqlashsa, u holda {\displaystyle \sum \_{n}b\_{n}} ham uzoqlashadi.

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

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

{\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
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Bir tomonlama taqqoslash alomati (testi) yuqori limitni qo'llagan holda keltirilishi mumkin. Barcha {\displaystyle n} lar uchun {\displaystyle a\_{n},b\_{n}\geq 0} bo'lsin. U holda agar {\displaystyle \limsup \_{n\to \infty }{\frac {a\_{n}}{b\_{n}}}=c} munosabat {\displaystyle 0\leq c<\infty } uchun o'rinli bo'lsa va {\displaystyle \Sigma \_{n}b\_{n}} qator yaqinlashuvchi bo'lsa, u holda zaruriy ravishda {\displaystyle \Sigma \_{n}a\_{n}} qator ham yaqinlashadi.

Misol
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Barcha natural {\displaystyle n} lar uchun {\displaystyle a\_{n}={\frac {1-(-1)^{n}}{n^{2}}}} va {\displaystyle b\_{n}={\frac {1}{n^{2}}}} bo'lsin. Ushbu{\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, {\displaystyle \limsup \_{n\to \infty }{\frac {a\_{n}}{b\_{n}}}=\limsup \_{n\to \infty }(1-(-1)^{n})=2\in [0,\infty )} munosabatdan va {\displaystyle \sum \_{n=1}^{\infty }{\frac {1}{n^{2}}}} ning yaqinlashuvchi ekanligidan, bir tomonlama taqqoslash alomatiga ko'ra {\displaystyle \sum \_{n=1}^{\infty }{\frac {1-(-1)^{n}}{n^{2}}}} qator yaqinlashuvchi bo'ladi.

Bir tomonlama taqqoslash alomatining teskarisi
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Barcha {\displaystyle n} lar uchun {\displaystyle a\_{n},b\_{n}\geq 0} bo'lsin. Agar {\displaystyle \Sigma \_{n}a\_{n}} qator uzoqlashuvchi bo'lib, {\displaystyle \Sigma \_{n}b\_{n}} qator yaqinlashuvchi bo'lsa, u holda zaruriy ravishda {\displaystyle \limsup \_{n\to \infty }{\frac {a\_{n}}{b\_{n}}}=\infty }, qaysiki, {\displaystyle \liminf \_{n\to \infty }{\frac {b\_{n}}{a\_{n}}}=0} munosabat o'rinli bo'ladi. Bu yerda asosiy mazmun, qaysidir ma'noda, {\displaystyle a\_{n}} larning {\displaystyle b\_{n}} lardan kattaroq ekanligidadir.

Misol
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{\displaystyle D=\{z\in \mathbb {C} :|z|<1\}} birlik diskda analitik bo'lgan{\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 {\displaystyle f} ning tasviri(aks)ning yuzasi {\displaystyle \sum \_{n=1}^{\infty }n|a\_{n}|^{2}} ga teng. Bundan tashqari, {\displaystyle \sum \_{n=1}^{\infty }1/n} qator uzoqlashuvchi. Shuning uchun, taqqoslash alomatining teskarisi bo'yicha,{\displaystyle \liminf \_{n\to \infty }{\frac {n|a\_{n}|^{2}}{1/n}}=\liminf \_{n\to \infty }(n|a\_{n}|)^{2}=0}, qaysiki, {\displaystyle \liminf \_{n\to \infty }n|a\_{n}|=0}.

Yana qarang
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* Yaqinlashish alomatlari
* Taqqoslash alomati (testi)

Manbalar
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1. 

1. ↑  Swokowski, Earl (1983), *Calculus with analytic geometry* (Alternate-nashr), Prindle, Weber & Schmidt, 516-bet, ISBN 0-87150-341-7

Qo'shimcha manbalar
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* Rinaldo B. Schinazi: *From Calculus to Analysis*. Springer, 2011, ISBN 9780817682897, 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
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* Pauls Online Notes on Comparison Test