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 $ \frac{1}{2}^{1}$ <= $\sum_{i=1}^{n} \frac{1}{2}^{n}$  < $\sum_{i=1}^{\infty} \frac{1}{2}^{n}$

We know, $\sum_{i=1}^{\infty} \frac{1}{2}^{n}$=$ \frac{1}{1-\frac{1}{2}}$ = 2

Hence we can say, $\sum_{i=1}^{n} \frac{1}{2}^{n}$ =O(1) [UpperBound]

and $\sum_{i=1}^{n} \frac{1}{2}^{n}$ = $\Omega$(1) [Lower Bound]

Therefore it is also $\Theta$(1)

 

 
 
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 $\sum_{i=1}^{1} \frac{1}{2}^{n}$ <= $\sum_{i=1}^{n} \frac{1}{2}^{n}$  < $\sum_{i=1}^{\infty} \frac{1}{2}^{n}$

 
 
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AfrikaansAlbanianAmharicArabicArmenianAzerbaijaniBasqueBelarusianBengaliBosnianBulgarianCatalanCebuanoChichewaChinese (Simplified)Chinese (Traditional)CorsicanCroatianCzechDanishDutchEnglishEsperantoEstonianFilipinoFinnishFrenchFrisianGalicianGeorgianGermanGreekGujaratiHaitian CreoleHausaHawaiianHebrewHindiHmongHungarianIcelandicIgboIndonesianIrishItalianJapaneseJavaneseKannadaKazakhKhmerKoreanKurdishKyrgyzLaoLatinLatvianLithuanianLuxembourgishMacedonianMalagasyMalayMalayalamMalteseMaoriMarathiMongolianMyanmar (Burmese)NepaliNorwegianPashtoPersianPolishPortuguesePunjabiRomanianRussianSamoanScots GaelicSerbianSesothoShonaSindhiSinhalaSlovakSlovenianSomaliSpanishSundaneseSwahiliSwedishTajikTamilTeluguThaiTurkishUkrainianUrduUzbekVietnameseWelshXhosaYiddishYorubaZulu
 
 
 
 
 
 
 
 
 
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