How to find the sum of the infinite geometric series given 1 + 2 3 + 4 9 + ... ?

Question

How to find the sum of the infinite geometric series given       1    +          2      3        +          4      9        +    ...  ?

Lola Rocha · Accepted Answer

Any geometric series&#039; general term can be expressed in the following way:            a      n        =    a    ⋅          r              n        -        1                                        for       n    =    1    ,    2    ,    3    ,    ...  where       a   is the initial term and       r   the common ratioIn our case we have:            a      n        =    1    ⋅                  (                  2          3                )                    n        -        1                                        for       n    =    1    ,    2    ,    3    ,    ...  with initial term       a    =    1   and common ratio       r    =          2      3      The general formula for the infinite sum (proved below) is:                    ∑                  n          =          1                ∞              a          r              n        -        1              =          a              1        -        r             when             |              r            |        &amp;lt;    1  So in our case:                    ∑                  n          =          1                ∞                            1              ⋅                  (                                            2              3                                      )                    n        -        1              =                            1                            1        -                                            2              3                                            =          1                        1          3                      =    3  BackgroundThe general term of a geometric series can be written:            a      n        =    a    ⋅          r              n        -        1                                        for       n    =    1    ,    2    ,    3    ,    ...  where       a   is the initial term and       r   is the common ratio.Given such a series, we find:            (      1      -      r      )                      ∑                  n          =          1                N              a          r              n        -        1              =                  ∑                  n          =          1                N              a          r              n        -        1              -    r                  ∑                  n          =          1                N              a          r              n        -        1                                          (          1          -          r          )                                      ∑                          n              =              1                        N                          a                  r                      n            -            1                                =                  ∑                  n          =          1                N              a          r              n        -        1              -    r                  ∑                  n          =          1                N              a          r              n        -        1                                          (          1          -          r          )                                      ∑                          n              =              1                        N                          a                  r                      n            -            1                                =                  ∑                  n          =          1                N              a          r              n        -        1              -                  ∑                  n          =          2                          N          +          1                      a          r              n        -        1                                          (          1          -          r          )                                      ∑                          n              =              1                        N                          a                  r                      n            -            1                                =    a    +                                                                                                            ∑                                          n                      =                      2                                        N                                                  a                                  r                                      n                    -                    1                                                                                            -                                                                                                            ∑                                          n                      =                      2                                        N                                                  a                                  r                                      n                    -                    1                                                                                            -    a          r      N                                    (          1          -          r          )                                      ∑                          n              =              1                        N                          a                  r                      n            -            1                                =    a    -    a          r      N                                    (          1          -          r          )                                      ∑                          n              =              1                        N                          a                  r                      n            -            1                                =    a          (      1      -              r        N            )      Dividing both ends by             (      1      -      r      )       we get the general finite sum formula:                                          ∑                          n              =              1                        N                          a                  r                      n            -            1                          =                              a                          (              1              -                              r                N                            )                                            1            -            r                              If             |              r            |        &amp;lt;    1   then             lim              N        →        ∞                    r      N        =    0   and we find:                    ∑                  n          =          1                ∞              a          r              n        -        1              =          lim              N        →        ∞                            ∑                  n          =          1                N              a          r              n        -        1                                                      ∑                          n              =              1                        ∞                          a                  r                      n            -            1                                =          lim              N        →        ∞                            a                  (          1          -                      r            N                    )                            1        -        r                                                      ∑                          n              =              1                        ∞                          a                  r                      n            -            1                                =          a              1        -        r            So we have the general formula for the infinite sum:                                          ∑                          n              =              1                        ∞                          a                  r                      n            -            1                          =                  a                      1            -            r                                                          when                               |                      r                    |                &amp;lt;        1

How to find the sum of the infinite geometric series given 1+2/3+4/9+...?

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New Questions in Calculus 2