Canthesumofallpositiveintegersbe−112Can\quad the\quad sum\quad of\quad all\quad positive\quad integers\quad be\quad -\frac { 1 }{ 12 } ?

Is∑1∞=−112?Is\quad \sum _{ 1 }^{ \infty } =-\frac { 1 }{ 12 } ?\\ \quad

#Algebra

Note by Kristian Vasilev
6 years, 5 months ago

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Comments

LetS=∑1∞,S1=1−1+1−1+1−1+...,S2=1−2+3−4+5−6+.....S1=1−1+1−1+1−1+....+S1=1−1+1−1+1−.....So2S1=1,thereforeS1=12,thenS2=1−2+3−4+5−6+....+S2=1−2+3−4+5−6+.....So2S2=1−1+1−1+1−1+...=12,thereforeS2=14S2=1−2+3−4+5−6+...=1+2+3+4+5+6+.....−2(2+4+6+8+...)S2=S−2×2S=−3S=14ThereforeS=−112.Let\quad S=\sum _{ 1 }^{ \infty } ,{ S }_{ 1 }=1-1+1-1+1-1+...,{ S }_{ 2 }=1-2+3-4+5-6+.....\\ \quad \quad \quad { S }_{ 1 }=1-1+1-1+1-1+....\quad \\ \quad \quad \quad +\\ \quad \quad \quad \quad \quad { S }_{ 1 }=1-1+1-1+1-.....\\ \quad So\quad 2{ S }_{ 1 }=1,therefore\quad { S }_{ 1 }=\frac { 1 }{ 2 } ,then\\ { S }_{ 2 }=1-2+3-4+5-6+....\\ +\\ \quad \quad { S }_{ 2 }=1-2+3-4+5-6+.....\\ So\quad 2{ S }_{ 2 }=1-1+1-1+1-1+...=\frac { 1 }{ 2 } ,therefore\quad { { S }_{ 2 }= }\frac { 1 }{ 4 } \\ { S }_{ 2 }=1-2+3-4+5-6+...=1+2+3+4+5+6+.....-2(2+4+6+8+...)\\ { S }_{ 2 }=S-2\times 2S=-3S=\frac { 1 }{ 4 } \\ Therefore\quad S=-\frac { 1 }{ 12 } .

Kristian Vasilev - 6 years, 5 months ago

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can you clarify summation of what? is it x

U Z - 6 years, 5 months ago

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Have I any mistakes?

Kristian Vasilev - 6 years, 5 months ago

I think this should be multiplication

Martin Nikolov - 6 years, 5 months ago

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@Martin Nikolov – yes it is

Kristian Vasilev - 6 years, 5 months ago
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