ከ«በር:ሒሳብ/ይህን ያውቁ ኖሯል?» ለውጦች መካከል ያለው ልዩነት

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መስመር፡ 1፦ መስመር፡ 1፦
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:<math>\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdot^{\cdot^{\cdot}}}}}}}}} = 2</math>
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መስመር፡ 12፦ መስመር፡ 11፦


:<math> \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots = e. </math> =2.71828.....
:<math> \frac{1}{0!} + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots = e. </math> =2.71828.....
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:<math> \frac{1}{1*2} + \frac{1}{3*4} + \frac{1}{5*6} + \frac{1}{7*8} + \cdots = ln{2}. </math> = 0.6931471.....
:<math> \frac{1}{1*2*3} + \frac{1}{4*5*6} + \frac{1}{7*8*9} + \frac{1}{9*10*11} + \cdots = \frac{1}{4}( \frac{\pi}{\sqrt{3}} - {ln{3}}). </math> = 0.178796.....
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: <math> \sqrt{1+2\sqrt{1+3 \sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+\cdots}}}}}} = 3. </math>
: <math> \sqrt{1+2\sqrt{1+3 \sqrt{1+4\sqrt{1+5\sqrt{1+6\sqrt{1+\cdots}}}}}} = 3. </math>
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:<math> 1+\cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2+ \cfrac{1}{2 + \cfrac{1}{2+{\ddots}}}}}} = \sqrt{2}.</math> = 1.41421...
:<math> 1+\cfrac{1}{2 + \cfrac{1}{2 + \cfrac{1}{2+ \cfrac{1}{2 + \cfrac{1}{2+{\ddots}}}}}} = \sqrt{2}.</math> = 1.41421...
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:<math>\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\sqrt{2}^{\cdot^{\cdot^{\cdot}}}}}}}}} = 2</math>

ነገር ግን!!

: <math> 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \cdots</math> ይህ ዝርዝር [[ሃርሞኒክ ዝርዝር]] ፣ ከ[[ምንም]] ቁጥር ጋር እኩል ነው፣ ምክንያቱም ያለማቋረጥ ድምር ውጤቱ እያደገ ስለሚሄድ።
የሚከተሉትን ጡቦች በዚህ ዝርዝር መሰረት ብንደረድራቸው፣ ማንኛውንም ክፍተት በድልድይ ማያያዝ እንችላለን፣ ከላይ እንደተጥቀሰው የጎን ርዝመቱ ሳያቋርጥ ስለሚጨምር (እስከ ዘላለም....)። ይሁንና ይህ ሃሳብ በመርህ ደረጃ እውነት ሆኖ በተግባር ግን እማይጠቅም ነው፣ ምክንያቱም የዝርዝሩ ድምር ውጤት እሚያድገው እጅግ ዝግ ብሎ ስለሆነ ነው።

[[File:Block_stacking_problem.svg|250px|center]]

እትም በ01:58, 10 ኦገስት 2019


=0.69314....
= 0.78539....
=1.64493....
=2.71828.....

= 0.6931471.....
= 0.178796.....


= 1.41421...