Double Summation

\[\begin{eqnarray} \sum _{ m=1 }^{ \infty }{ \sum _{ n=1 }^{ m }{ \frac { 1 }{ { 2 }^{ n+m } } } } & = \sum _{ n=1 }^{ \infty }{ \sum _{ m=n }^{ \infty }{ \frac { 1 }{ { 2 }^{ n } } \frac { 1 }{ { 2 }^{ m } } } } \\ \quad & = \sum _{ n=1 }^{ \infty }{ \sum _{ m=0 }^{ \infty }{ \frac { 1 }{ { 2 }^{ 2n } } \frac { 1 }{ { 2 }^{ m } } } } \\ \quad & = \sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 4^{ n } } \sum _{ m=0 }^{ \infty }{ \frac { 1 }{ { 2 }^{ m } } } } \\ \quad & = (\sum _{ m=0 }^{ \infty }{ \frac { 1 }{ { 2 }^{ m } } } )(\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ 4^{ n } } } ) \\ \quad & = (2)(\frac { 1 }{ 3 } ) \\ \quad & = \boxed { \frac { 2 }{ 3 } } \end{eqnarray}\]

#Algebra

Note by Gandoff Tan
2 years ago

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