Real number in terms of infinite series A = a 0 + ( 1 a 1 ) k + ( 1 a 2 ) k + ( 1 a 3 ) k + …Where i ≥ 1 and the recurrence relation a i + 1 ≥ a i ≥ 2

Question

Real number in terms of infinite series  A  =      a    0    +            (              1                  a          1                    )        k    +            (              1                  a          2                    )        k    +            (              1                  a          3                    )        k    +  …Where   i  ≥  1 and the recurrence relation       a          i      +      1        ≥      a    i    ≥  2

crociandomh · Accepted Answer

Given that from your condition it is allowed that       a          i      +      1        =      a    i  , if k is integer, you can reduce the problem for   k  &amp;gt;  1 to the case   k  =  1 by repeating   1      /        a    i    k   for       a    i          k      −      1       times before using a new value. By doing that repetition, you get a sum of       a    i          k      −      1            /        a    i    k    =  1      /        a    i  Example:Say your   k  =  1 expansion is  A  =  1  +  1      /    2  +  1      /    8  +  …and   k  =  2. Then your new expansion reads                    A                            =        1        +                              (                          1              2                        )                    2                +                              (                          1              2                        )                    2                                                                =            1                          +                              (                          1              8                        )                    2                +                              (                          1              8                        )                    2                +                              (                          1              8                        )                    2                +                              (                          1              8                        )                    2                                                                =            1                          +                              (                          1              8                        )                    2                +                              (                          1              8                        )                    2                +                              (                          1              8                        )                    2                +                              (                          1              8                        )                    2                +        …            Since every real number can be written as the product of an integer and a real number in the range   [  1  ,  2  ), the same strategy can be used to reduce the problem for arbitrary non-integer   k  &amp;gt;  1 to the problem   1  &amp;lt;  k  &amp;lt;  2

Real number in terms of infinite series A = a 0 </msub> + (

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New Questions in Calculus 2