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Question Number 143769 by bemath last updated on 18/Jun/21

	Answered by TheHoneyCat last updated on 18/Jun/21
	$(1/2)+(1/6)+(1/(12))+(1/(20))+(1/(9900))=((7921)/(990))≈0.8001010101 IF ′′+...+′′, in your question, is a finite sum of positive real numbers: (1) (9/(100))=0.09<0.8 ⇒ (A) is false ((19)/(100))=0.19<0.8 ⇒ (C) is false ((99)/(200))=0.495<0.8 ⇒ (D) is false ((99)/(1000))=>0.099<0.8 ⇒ (E) is false IF the right result is necessarly among {(A),(B),(C),(D),(E)} (2) well the answere is (B) however I don′t see why these two hypothesis are true. usualy ′′+...+′′ refers to an obvious sequence like, for instance, (1/2)+(1/6)+(1/(24))+... is (1/(n!)) and (1/2)+(1/6)+(1/(30))+(1/(210))+...=(1/(Π_(k=1) ^n p_n )) where p_n is the n^(th) prime number in this case, I don′t know what +...+ refers too. so obviously my proof is a bit lame$
	$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + \frac{1}{9900} = \frac{7921}{990} \approx 0.8001010101$ $IF^{″} + . . . +^{″}, in your question, is a finite sum of positive real numbers : (1)$ $\frac{9}{100} = 0.09 < 0.8 \Rightarrow (A) is false$ $\frac{19}{100} = 0.19 < 0.8 \Rightarrow (C) is false$ $\frac{99}{200} = 0.495 < 0.8 \Rightarrow (D) is false$ $\frac{99}{1000} => 0.099 < 0.8 \Rightarrow (E) is false$ $IF the right result is necessarly among {(A), (B), (C), (D), (E)} (2)$ $well the answere is (B)$ $h o w e v e r I {d o n}^{'} t s e e w h y t h e s e t w o h y p o t h e s i s a r e t r u e .$ $u s u a l y^{″} + . . . +^{″} r e f e r s t o a n o b v i o u s s e q u e n c e$ $l i k e, f o r i n s t a n c e, \frac{1}{2} + \frac{1}{6} + \frac{1}{24} + . . . i s \frac{1}{n!}$ $a n d \frac{1}{2} + \frac{1}{6} + \frac{1}{30} + \frac{1}{210} + . . . = \frac{1}{\underset{k = 1}{\prod^{n}} p_{n}} w h e r e p_{n} i s t h e n^{t h} p r i m e n u m b e r$ $i n t h i s c a s e, I {d o n}^{'} t k n o w w h a t + . . . + r e f e r s t o o .$ $s o o b v i o u s l y m y p r o o f i s a b i t l a m e$
	Answered by Dwaipayan Shikari last updated on 18/Jun/21

	$\frac{1}{2} + \frac{1}{6} + \frac{1}{12} + \frac{1}{20} + . . + \frac{1}{9900}$ $= 1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{3} + \frac{1}{3} - \frac{1}{4} + . . \frac{1}{99} - \frac{1}{100}$ $= 1 - \frac{1}{100} = \frac{99}{100}$
	Commented by Dwaipayan Shikari last updated on 18/Jun/21

	$G e n r e a l C a s e f o r a p p r o x i m a t i n g$ $\underset{k = 1}{\sum^{n}} \frac{1}{k} \approx γ + l o g (n)$
	Answered by MJS_new last updated on 18/Jun/21

	$\frac{1}{2} + \frac{1}{2 + 4} + \frac{1}{2 + 4 + 6} + \frac{1}{2 + 4 + 6 + 8} + . . .$ $= \underset{k = 1}{\sum^{n}} \frac{1}{k^{2} + k} = \frac{n}{n + 1}$ $in this case n = 99 \Rightarrow answer is \frac{99}{100}$
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