For any positive integer n , let n ^ denote the integer nearest to n . Then how to prove that ∑ m = 1 ∞ 2 m ^ + 2 − m ^ 2 m = 3

Question

For any positive integer n , let             n      ^       denote the integer nearest to       n  . Then how to prove that      ∑          m      =      1        ∞                      2                                            m              ^                                          +              2                  −                                    m              ^                                                  2              m              =  3

Regan Holloway · Accepted Answer

Let   f  (  m  )  =  m  −            m      ^        ,  g  (  m  )  =  m  +            m      ^      , then      ∑          m      =      1              ∞                                    2                                                    m                ^                                                    +                  2                      −                                          m                ^                                                                2                  m                      =      ∑          m      =      1              ∞                  (                        (                      1            2                    )                          m          −                                    m              ^                                          +                        (                      1            2                    )                          m          +                                    m              ^                                          )        =      ∑          m      =      1              ∞                          (                  1          2                )                    f        (        m        )              +      ∑          m      =      1              ∞                          (                  1          2                )                    g        (        m        )            Note that                    m        +        1            ^        −            m      ^        =      {                            1                          if                    m          =                      k            2                    +          k          ,                    k          ∈                                    Z                        +                                                0                          otherwise                        Thus  f  (  m  +  1  )  −  f  (  m  )  =  1  −  (                    m        +        1            ^        −            m      ^        )  =      {                            0                          if                    m          =                      k            2                    +          k          ,                    k          ∈                                    Z                        +                                                1                          otherwise                        Then                                            ∑                      m            =            1                                ∞                                                              (                              1                2                            )                                      f              (              m              )                                      +                  ∑                      m            =            1                                ∞                                                              (                              1                2                            )                                      g              (              m              )                                                                        =                  (                      ∑                          m              =              0                                      ∞                                                                          (                                  1                  2                                )                            m                                +                      ∑                          m              =                              k                2                            ,              k              ≥              1                                                                          (                                  1                  2                                )                            m                                )                +                  (                      ∑                          m              =              2                                      ∞                                                                          (                                  1                  2                                )                            m                                −                      ∑                          m              =                              k                2                            ,              k              ≥              2                                                                          (                                  1                  2                                )                            m                                )                                                  =                              (                          1              2                        )                    0                +        2                  ∑                      m            =            1                                ∞                                                              (                              1                2                            )                        m                                                            =        3

For any positive integer n , let nˆ denote the integer nearest to sqrt(n). Then how to prove that sum_(m=1)^infty (2^m+2^(-m))/2^m=3

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