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\(Can\quad the\quad sum\quad of\quad all\quad positive\quad integers\quad be\quad -\frac { 1 }{ 12 }\) ?

\(Is\quad \sum _{ 1 }^{ \infty } =-\frac { 1 }{ 12 } ?\\ \quad\)

Note by Kristian Vasilev
2 years, 7 months ago

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\(Let\quad S=\sum _{ 1 }^{ \infty } ,{ S }_{ 1 }=1-1+1-1+1-1+...,{ S }_{ 2 }=1-2+3-4+5-6+.....\\ \quad \quad \quad { S }_{ 1 }=1-1+1-1+1-1+....\quad \\ \quad \quad \quad +\\ \quad \quad \quad \quad \quad { S }_{ 1 }=1-1+1-1+1-.....\\ \quad So\quad 2{ S }_{ 1 }=1,therefore\quad { S }_{ 1 }=\frac { 1 }{ 2 } ,then\\ { S }_{ 2 }=1-2+3-4+5-6+....\\ +\\ \quad \quad { S }_{ 2 }=1-2+3-4+5-6+.....\\ So\quad 2{ S }_{ 2 }=1-1+1-1+1-1+...=\frac { 1 }{ 2 } ,therefore\quad { { S }_{ 2 }= }\frac { 1 }{ 4 } \\ { S }_{ 2 }=1-2+3-4+5-6+...=1+2+3+4+5+6+.....-2(2+4+6+8+...)\\ { S }_{ 2 }=S-2\times 2S=-3S=\frac { 1 }{ 4 } \\ Therefore\quad S=-\frac { 1 }{ 12 } .\) Kristian Vasilev · 2 years, 7 months ago

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@Kristian Vasilev can you clarify summation of what? is it x Megh Choksi · 2 years, 7 months ago

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@Megh Choksi Have I any mistakes? Kristian Vasilev · 2 years, 7 months ago

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@Megh Choksi I think this should be multiplication Martin Nikolov · 2 years, 7 months ago

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@Martin Nikolov yes it is Kristian Vasilev · 2 years, 7 months ago

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